Question 513 Marks
Show graphically how gravitational field strength varies with distance from the centre of earth, outwards. Give the relation also.
AnswerFor points inside the earth, $\text{E}_{\text{gi}}=\frac{\text{G}\frac{4}{3}\pi\text{x}^3}{\text{x}^2}=\frac{4\text{G}}{3}\pi\text{x}\rho$ 
For points outside the earth, $\text{E}_{\text{go}}=\frac{\text{G}\frac{4}{3}\pi\text{R}^3\rho}{\text{x}^2}=\frac{\text{GM}}{\text{x}^2}$ View full question & answer→Question 523 Marks
A sphere of mass $40kg$ is attracted by another mass of $15kg$ when their centres are $20cm$ apart, with a force of $1 \times 10^{-6} N$. Calculate the value of gravitational constant.
AnswerGiven: $\text{m}=40\text{kg}$
$\text{m}_2=15\text{kg}$
$\text{r}=20\text{cm}=0.2\text{m}$
$\text{F}=1\times10^{-6}\text{N}$
$\text{G}=?$
$\text{F}=\text{G}\frac{\text{m}_1\text{m}_2}{\text{r}^2}$
$1\times10^{-6}=\text{G}\ \times\frac{40\times15}{(0.2)^2}$
$\text{G}=\frac{1\times10^{-6}\times(0.2)^2}{15\times140}$
$=\frac{0.04\times10^{-6}}{600}$
$=6.66\times10^{-11}\ \text{Nm}^2/\text{Kg}^2$
View full question & answer→Question 533 Marks
The planet Saturn has a mass $95$ times that of the earth, and its radius is $9.5$ times the earth's radius. Calculate the escape speed of a body from Saturn's surface, if the escape speed from the earth's surface is $11.2kms^{-1}$.
AnswerEscape speed from the earth's surface is, $\text{v}_\text{e}=\Big[\frac{2\text{GM}}{\text{R}}\Big]^\frac{1}{2}$ or $\Big[\frac{2\text{GM}}{\text{R}}\Big]^\frac{1}{2}=11.2$ Escape speed from Saturn's surface will be,$\text{v}=\Big[\frac{2\text{GM}'}{\text{R}'}\Big]^\frac{1}{2}$
Now, $\text{M}'=95\text{M},\text{ R}'=9.5\text{R}$
$\text{v}=\Big[\frac{2\times95\text{GM}}{9.5\text{R}}\Big]^\frac{1}{2}=3.16\times\sqrt{2\frac{\text{GM}}{\text{R}}}$
But $\Big[\frac{2\text{GM}}{\text{R}}\Big]^\frac{1}{2}=11.2\text{km s}^{-1}$
$\therefore\ \text{v}=3.16\times11.2=35.4\text{km s}^{-1}.$
View full question & answer→Question 543 Marks
Find an expression for the orbital velocity of a satellite revolving around the earth in a circular orbit at a height h above the surface of earth.
AnswerConsider a satellite of mass m revolving around the earth at a height h from its surface so that radius of its orbit r = R + h. If $v_0$ be the orbital velocity of satellite then centripetal force needed by it for its uniform circular motion is, $\text{F}=\frac{\text{mv}_0^2}{\text{r}}$ This value of centripetal force is provided by the gravitational pull of the earth acting on the satellite i.e., $\text{F}=\frac{\text{GMm}}{\text{r}^2}$ For equilibrium, $\frac{\text{mv}^2_0}{\text{r}}=\frac{\text{GMm}}{\text{r}^2}$$\Rightarrow\ \text{v}_0=\sqrt{\frac{\text{GM}}{\text{r}}}=\sqrt{\frac{\text{GM}}{(\text{R}+\text{h})}}$
But $\text{g}=\frac{\text{GM}}{\text{R}^2},$ hence $GM = gR^2$
$\therefore\ \text{v}_0=\sqrt{\frac{\text{gR}^2}{(\text{R}+\text{h})}}=\text{R}\sqrt{\frac{\text{g}}{(\text{R}+\text{h})}}.$
View full question & answer→Question 553 Marks
Satellite $A$ is in a certain circular orbit about a planet, while satellite $B$ is in a larger circular orbit. Which satellite has $(i)$ the longer period and $(ii)$ the greater speed?
Answer
- $\text{T}=2\pi\sqrt{\frac{\text{r}}{\text{GM}}}$
$\therefore$ larger $'r\ '$,larger $T$
$\therefore$ Satellite B has longer period.
- $\text{v}_{\text{o}}=\sqrt{\frac{\text{GM}}{\text{r}}},$ lesser $r, $ more $v_o$
$\therefore$ Satellite $A$ has greater speed. View full question & answer→Question 563 Marks
While approaching a planet circling a distant star, a space traveller determines the planet's radius to be half that of the earth. After landing on the surface, he finds the acceleration due to gravity to be twice that on the surface of the earth. Find the ratio of the mass of the planet to that of the earth.
AnswerIn case of the earth, $\frac{\text{GM}_{\text{e}}\text{m}}{\text{r}^2_{\text{e}}}=\text{mg}_{\text{e}}$ In case of the planet, $\frac{\text{GM}_{\text{p}}\text{m}}{\text{r}^2_{\text{p}}}=\text{mg}_{\text{p}}$ Dividing these two equations, we get $\Big(\frac{\text{M}_{\text{p}}}{\text{M}_{\text{e}}}\Big)\Big(\frac{\text{r}^2_{\text{e}}}{\text{r}^2_{\text{p}}}\Big)=\frac{\text{g}_{\text{p}}}{\text{g}_{\text{e}}},$ $\text{but }\text{g}_{\text{p}}=2\text{g}_{\text{e}}\text{ and }\text{r}_{\text{p}}=\frac{\text{r}_{\text{e}}}{2}$ $\therefore\frac{\text{M}_{\text{p}}}{\text{M}_{\text{e}}}=\frac{2}{4}=\frac{1}{2}$ Thus the ratio of the mass of the planet to the mass of the earth is $\frac{1}{2}.$
View full question & answer→Question 573 Marks
The mass of a spaceship is $1000kg$. It is to be launched from the earth's surface out into free space. The value of g and R (radius of earth) are $10m/ s^2$ and $6400km$, respectively. What is the required energy for this work done?
Answer$\text{W}=0-\Big[\frac{-\text{GMm}}{\text{R}}\Big]=\frac{\text{GMm}}{\text{R}}$
$=\text{gR}^2\times\frac{\text{m}}{\text{R}}=\text{mgR}$
$=1000\times10\times6400\times10^3$
$=64\times10^9\text{J}=6.4\times10^{10}\text{J}$
View full question & answer→Question 583 Marks
Derive an expression for work done against gravity.
AnswerPotential energy of the body on the surface of the earth $=\frac{-\text{GMm}}{\text{R}}$ Potential energy of the body at a heighth from the surface of the earth $=-\frac{\text{GMm}}{(\text{R}+\text{h})}$ Work done $=\Big(-\frac{\text{GMm}}{\text{R}+\text{h}}\Big)-\Big(-\frac{\text{GMm}}{\text{R}}\Big)$ $=\frac{\text{GMm}}{\text{R}}-\frac{\text{GMm}}{\text{R}+\text{h}}$ $=\text{GMm}\Big(\frac{1}{\text{R}}-\frac{1}{\text{R}+\text{h}}\Big)$ $=\frac{\text{GMm}}{\text{R}}-\frac{\text{GMm}}{\text{R}+\text{h}}$ $=\text{GMm}\Big(\frac{1}{\text{R}}-\frac{1}{\text{R}+\text{h}}\Big)$ $=\frac{\text{GMmh}}{\text{R}(\text{R}+\text{h})}=\frac{\text{MgR}^2\text{h}}{\text{R}(\text{R}+\text{h})}$ $\Big[\because\text{g}=\frac{\text{GM}}{\text{R}^2}\Big]$ $=\frac{(\text{Mgh})}{(\text{R}+\text{h})}=\frac{\text{Mgh}}{1+\frac{\text{h}}{\text{R}}}$
View full question & answer→Question 593 Marks
A body is released at a distance r from the centre of the earth. Prove that the velocity v of the body when it strikes the surface of the earth is given by, $\text{v}=\text{R}\Big[2\text{g}\Big(\frac{1}{\text{R}}-\frac{1}{\text{r}}\Big)\Big]^{\frac{1}{2}}$ where R is the radius of the earth r > R.
AnswerP.E. at A = (P.E. + K.E.) at B 
$-\frac{\text{GMm}}{\text{r}}=-\frac{\text{GMm}}{\text{R}}+\frac{1}{2}\text{mv}^2$ $\text{v}^2=\frac{2\text{GM}}{\text{R}}-\frac{2\text{GM}}{\text{R}}$ $=\frac{2\text{GMR}}{\text{R}^2}-\frac{2\text{GM}}{\text{R}^2}\frac{\text{R}^2}{\text{r}}$ $=2\text{gR}\big(1-\frac{\text{R}}{\text{r}}\Big)$ $\text{v}=\sqrt{2\text{gR}\Big(1-\frac{\text{R}}{\text{r}}\Big)}$ $=\sqrt{2\text{gR}^2\Big(\frac{1}{\text{R}}-\frac{1}{\text{r}}\Big)}$ $=\text{R}\sqrt{2\text{g}\Big(\frac{1}{\text{R}}-\frac{1}{\text{r}}\Big)}$ View full question & answer→Question 603 Marks
What is gravitational potential energy at a point? How much of work is done in shifting a mass from the surface to a height equal to its radius?
AnswerGravitational potential energy is the work done in shifting a mass m from one point to the other. At any distance x from the centre of earth (M), the gravitational force is, $\frac{\text{GMm}}{\text{x}^2}$ $\therefore$ Work done in shifting (m) from the surface to a height equal to the radius, then, $\text{W}=\int^\limits{2\text{R}}_\limits{\text{R}}\frac{\text{GMm}}{\text{x}^2}\text{dx}$$=\text{GMm}\Big[-\frac{1}{\text{x}}\Big]^{2\text{R}}_{\text{R}}=-\frac{\text{GMm}}{\text{2R}}$
View full question & answer→Question 613 Marks
Two planets have masses in the ratio $1 : 10$ and radii in the ratio $2 : 5$. Compare:
- Their densities.
- The acceleration due to gravity on their surface.
- Escape velocities from their surfaces.
- The periods of revolutions of satellites near to their surfaces.
AnswerLet $M_1, M_2$ by the masses and $R_1, R_2$ be the radii of the planets.
$\Rightarrow \frac{\text{M}_1}{\text{M}_2}=\frac{1}{10}$ and $\frac{\text{R}_1}{\text{R}_2}=\frac{2}{5}$
- Ratio of densities $=\frac{\text{d}_1}{\text{d}_2}$
$\frac{\text{d}_1}{\text{d}_2}=\bigg[\frac{\text{M}_1}{\frac{4}{3}\pi\text{R}_1^3}\bigg]\bigg[\frac{\frac{4}{3}\pi\text{R}^3_2}{\text{M}_2}\bigg]$
$\frac{\text{d}_1}{\text{d}_2}=\frac{\text{M}_1}{\text{M}_2}\Big[\frac{\text{R}_2}{\text{R}_1}\Big]^3$
$\frac{\text{d}_1}{\text{d}_2}=\Big[\frac{1}{10}\Big]\Big[\frac{5}{2}\Big]^3=\frac{25}{16}$
- Acceleration due to gravity at the surface $=\text{g}=\frac{\text{GM}}{\text{R}^2}$
$\therefore\ \frac{\text{g}_1}{\text{g}_2}=\frac{\text{M}_1}{\text{M}_2}\Big[\frac{\text{R}_2}{\text{R}_1}\Big]^2$
$=\frac{1}{10}\Big[\frac{5}{2}\Big]^2=\frac{5}{8}$
- Escape velocity $=\sqrt{\frac{2\text{GM}}{\text{R}}}$
$\Rightarrow\frac{\text{v}_1}{\text{v}_2}=\sqrt{\frac{\text{M}_1}{\text{M}_2}}\sqrt{\frac{\text{R}_2}{\text{R}_1}}=\sqrt{\frac{1}{10}\times\frac{5}{2}}=\frac{1}{2}$
- Time period of a satellite near the surface $($orbit radius $= R) =\frac{2\pi}{\sqrt{\text{GM}}}\text{R}\sqrt{\text{R}}$
$\Rightarrow\frac{\text{T}_1}{\text{T}_2}=\sqrt{\frac{\text{M}_2}{\text{M}_1}}\Big[\frac{\text{R}_1}{\text{R}_2}\Big]\Big[\sqrt{\frac{\text{R}_1}{\text{R}_2}}\Big]$
$=\sqrt{\frac{10}{1}}\Big[\frac{2}{5}\Big]\Big[\sqrt{\frac{2}{5}}\Big]=\frac{4}{5}$ View full question & answer→Question 623 Marks
- According to Kepler's second law, the radius vector to a planet from the sun sweeps out equal areas in equal interval of time. The law is consequence of which conservation law?
- State Kepler's third law.
Answer
- Law of conservation of angular momentum.
- Kepler's third law is also known as law of periods, the square of the period of revolution of a planet around the Sun is proportional to the cube of semi major axis of elliptical orbit.
View full question & answer→Question 633 Marks
The change in the value of g at a height h above the earth is same as at a depth d below it. If h and d are small as compared to the radius of the earth, what is the ratio $\Big(\frac{\text{h}}{\text{d}}\Big)?$
Answer$\text{g}_{\text{h}}=\frac{\text{gR}^2}{(\text{R}+\text{h})^2}=\frac{\text{gR}^2}{\text{R}^2\Big(1+\frac{\text{h}}{\text{R}}\Big)^2}$ $=\text{g}\Big(1+\frac{\text{h}}{\text{R}}\Big){-2}=\text{g}\Big(1-\frac{2\text{h}}{\text{R}}\Big)$ $\text{g}_{\text{d}}=\text{g}\Big(1-\frac{\text{d}}{\text{R}}\Big)\dots(2)$ $\text{But }\text{g}\Big(1-\frac{2\text{h}}{\text{R}}\Big)=\text{g}\Big(1-\frac{\text{d}}{\text{R}}\Big)$ $\text{ or }1-\frac{2\text{h}}{\text{R}}=1-\frac{\text{d}}{\text{R}}\text{ or }\frac{\text{h}}{\text{d}}=\frac{1}{2}$
View full question & answer→Question 643 Marks
If the earth has a mass $9$ times and radius twice of the planet Mars, calculate the minimum speed required by a rocket to pull out of the gravitational force of Mars. Escape speed on the surface of the earth is $11.2km s^{-1}$.
AnswerEscape speed on the surface of earth is, $\text{v}_\text{e}=\sqrt{\frac{2\text{GM}}{\text{R}}}=11.2\text{km s}^{-1}$ Now, Mass of Mars $=\frac{\text{M}}{9}$ Radius of Mars $=\frac{\text{R}}{2}$ $\therefore$ Escape speed on the surface of Mars is,$\text{v}_\text{m}=\sqrt{\frac{2\text{G}\big(\frac{\text{M}}{9}\big)}{\frac{\text{R}}{2}}}$
$=\sqrt{\frac{4}{9}\frac{\text{GM}}{\text{R}}}=\frac{\sqrt{2}}{3}\times\text{v}_\text{e}$
$=\frac{1.414}{3}\times11.2\text{km s}^{-1}=5.279\text{km s}^{-1}$
View full question & answer→Question 653 Marks
What is escape velocity. Derive an expression for the same.
AnswerThe minimum velocity required to escape from the gravitational force of earth is called escape velocity. Total energy is the sum of P.E. and K.E. $\text{T.E}=-\frac{\text{GMm}}{\text{R}}+\frac{1}{2}\text{mv}^2$ To escape K.E should be greater than P.E., i.e.. $\frac{1}{2}\text{mv}^2\geq-\frac{\text{GMm}}{\text{R}}$ $\text{v}_{\text{e}}=\sqrt{2\frac{\text{GM}}{\text{R}}}=\sqrt{2\text{gR}}$
View full question & answer→Question 663 Marks
A projectile is fired vertically upward from the surface of earth with a speed $kv_e$, where $v_e$. is the escape speed and $k < 1$. Neglecting air resistance show that the maximum height to which it will rise measured from the centre of earth is $\frac{\text{R}}{(1-\text{k}^2)},$ where R is the radius of the earth.
AnswerLet a body of mass m be projected from the surface of earth with speed v and it reaches to a height h. Using law of conservation of energy (relative to surface of earth) we have, $\frac{1}{2}\text{mv}^2=\frac{\text{mgh}}{1+\frac{\text{h}}{\text{R}}}$ In this problem, $\text{v}=\text{kv}_\text{e}=\text{k}\sqrt{2\text{gR}}$ and h = r - R So, $\frac{1}{2}\text{mk}^22\text{gR}=\frac{\text{mg(r}-\text{R})}{\Big[1+\frac{(\text{r}-\text{R})}{\text{R}}\Big]}$
$\text{k}^2=\frac{\text{r}-\text{R}}{\text{r}}=1-\frac{\text{R}}{\text{r}}$
$\Rightarrow\ \text{r}=\frac{\text{R}}{1-\text{k}^2}.$
View full question & answer→Question 673 Marks
Out of aphelion and perihelion, where is the speed of the earth more and why?
AnswerThe earth revolves around the sun in an elliptical orbit or by Kepler’s first law and sun remains at its one focus. The position of earth at P and A at shortest and longest distance are called perihelion and Aphelion respectively. 
According to the second law of kepler’s the areal velocity of planet around the sun is constant. $\frac{\text{dA}}{\text{dt}}=\frac{\text{L}}{2\text{m}}=\frac{\text{r}\times\text{p}}{2\text{m}}=\frac{\text{r}\times\text{mv}}{2\text{m}}=\frac{1}{2}\text{r}\times\text{v}$ Hence, if r increases at Aphelion the v decreases and vice - versa at p. View full question & answer→Question 683 Marks
Define gravitational potential and gravitational field intensity. Give their units. Also write the relation between them.
AnswerGravitational potential at a point in gravitational field of the body is defined as the amount of work done in bringing a body of unit mass from infinity to that point without acceleration. It has units of J/ kg. Gravitational field intensity is defined as the force experienced by a body of unit mass placed at that point provided the presence of unit mass does not disturb the original gravitational field. It is expressed in N/ kg. $\text{I}=\frac{\text{V}_{\text{p}}}{\text{r}}$
View full question & answer→Question 693 Marks
An astronaut, on his journey between the earth and the moon experience weightlessness at a point $P$. Find the distance of the point $P$ from the centre of the earth. (Given: mass of earth is 80 times the mass of the moon and the distance between them is $3.84 \times 10^5 \mathrm{~km}$ ).
AnswerLet the distance between P and the centre of earth be x. At this point the gravitational pull due to earth on a mass m balances the gravitational pull due to the moon. $\text{F}=\text{G}\frac{\text{M}_{\text{E}}\text{m}}{(\text{d}-\text{x})^2}$
$\text{or }\frac{\text{M}_{\text{E}}}{\text{x}^2}=\frac{\text{M}_{\text{m}}}{(\text{d}-\text{x})^2}$
$\frac{(\text{d}-\text{x})^2}{\text{x}^2}=\frac{\text{M}_{\text{m}}}{\text{M}_{\text{E}}}$
$\frac{\text{d}-\text{x}}{\text{x}}=\frac{1}{\sqrt{80}}$
$\frac{\text{d}-\text{x}}{\text{x}}=\frac{1}{8.94}$
$\text{x}=8.94\text{d}-8.94\text{x}$
$9.94\text{x}=8.94\text{d}$
$\text{x}=\frac{8.94\times3.84\times10^5}{9.94}$
$(\dots\text{d}=3.84\times10^5\text{km})$
$=3.4\times10^5\text{km}.$
View full question & answer→Question 703 Marks
The acceleration due to gravity on the moon is only one-sixth of that on earth. Suppose the average density of both are same, what would be the ratio of the radii of the moon and the earth?
Answer$\text{g}=\frac{\text{GM}}{\text{R}^2}$ $\frac{1}{6}=\frac{\text{g}_{\text{moon}}}{\text{g}_{earth}}=\frac{\text{R}^2_{\text{e}}\text{M}_{\text{m}}}{\text{R}^2_{\text{m}}\text{M}_{\text{e}}}$ $=\frac{\text{R}^2_{\text{e}}}{\text{R}^2_{\text{m}}}\times\frac{\frac{4}{3}\pi\text{R}^3_{\text{m}}\rho}{\frac{4}{3}\pi\text{R}^3_{\text{e}}\rho}=\frac{\text{R}_{\text{m}}}{\text{R}_{\text{e}}}$ $\therefore\text{R}_{3}=\frac{1}{6}\text{R}_{\text{e}}.$
View full question & answer→Question 713 Marks
Taking the moon's orbit around earth to be r and mass of earth 81 times the mass of the moon. Find the position of the point from the earth, where the net gravitational field is zero.
AnswerLet x be the distance of a point from the earth where resultant gravitational field intensity is zero. So, $\frac{\text{GM}_\text{e}}{\text{x}^2}=\frac{\text{GM}_\text{e}}{(\text{r}-\text{x})^2}$ $\frac{81\text{M}_\text{m}}{\text{x}^2}=\frac{\text{M}_\text{m}}{(\text{r}-\text{x})^2}$ $\frac{9}{\text{x}}=\frac{1}{(\text{r}-\text{x})}$ $9\text{r}=10\text{x}$ $\text{x}=\frac{9\text{r}}{10}=0.9\text{r}$
View full question & answer→Question 723 Marks
What is the direction of areal velocity of the earth around the sun?
AnswerAreal velocity of the earth around the sun is given by $\frac{\vec{\text{dA}}}{\text{dt}}=\frac{\vec{\text{L}}}{2\text{M}}$ 
where, L is the angular momentum and M is the mass of the earth. But angular momentum $\vec{\text{L}}=\vec{\text{r}}\times\vec{\text{p}}=\vec{\text{r}}\times\text{m}\vec{\text{v}}$ Therefore, the direction of a real velocity$\Big(\frac{\vec{\text{dA}}}{\text{dt}}\Big)$is in the direction of $(\vec{\text{r}}\times\vec{\text{v}}),$ i.e., perpendicular to the plane containing r and v and directed as given by right hand rule. So, areal velocity is normal to the plane containing Earth and Sun as shown in the figure. View full question & answer→Question 733 Marks
The mass and diameter of a planet are twice of those of the earth. What will be the period of oscillation of a pendulum on this planet, if it is a second's pendulum on the earth?
AnswerWe know, $\text{g}=\frac{\text{GM}_{\text{e}}}{\text{R}^2}$
$\therefore\text{g}_{\text{e}}=\frac{\text{GM}_{\text{e}}}{\text{R}^2_{\text{e}}}$ $\text{and }\text{g}_{\text{p}}=\frac{\text{GM}_{\text{p}}}{\text{R}^2_{\text{p}}}$ Given : $M_p = 2M_e and R_p = 2Re$
$\therefore\frac{\text{g}_{\text{p}}}{\text{g}_{\text{e}}}=\frac{1}{2}$
The time period of a simple pendulum is given by $\text{T}_{\text{e}}=2\pi\sqrt{\frac{\text{l}}{\text{g}_{\text{e}}}}$ $\text{and }\text{T}_{\text{p}}=2\pi\sqrt{\frac{\text{l}}{\text{g}_{\text{p}}}}$ $\text{For }\text{T}_{\text{e}}=1\text{s},\text{T}_{\text{p}}=\sqrt{2}\text{s}.$
View full question & answer→Question 743 Marks
A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun $= 2 \times 10^{30}kg$, mass of the earth $= 6\times 10^{24}kg$. Neglect the effect of other planets etc. (orbital radius $= 1.5 \times 10^{11}m).$
AnswerGiven: Mass of Sun $M = 2 \times 10^{30}kg$ Mass of the earth $m = 6 \times 10^{24}kg$
Distance between Sun and Earth $r = 1.5 \times 10^{11}m$ Consider a point R,
where the gravitational force on the rocket due to earth = gravitational force on the rocket due to sun.
Distance of point R from the Earth = x
Therefore, $\frac{(\text{Gm})}{(\text{r}^2)}=\frac{(\text{GM})}{(\text{r}-\text{x})^2}$
$\frac{(\text{r}-\text{x})^2}{(\text{x})^2}=\Big(\frac{\text{M}}{\text{m}}\Big)$
$=\frac{(2\times10^{30})}{(6\times10^{24})}$
$=\frac{(10^{6})}{(3)}$
$=\frac{(\text{r}-\text{x})}{(\text{x})}=\frac{(10)^3}{(\sqrt{3})}$
$\big(\frac{\text{r}}{\text{x}}\big)=\frac{(10)^3}{(\sqrt{3})+1}$
$\approx\frac{(10)^3}{(\sqrt{3})}$
$\text{x}=\frac{(\sqrt{3})\text{r}}{(10)^3}$
$=\frac{(1.732\times1.5\times10^{11})}{(10^3)}\text{m}$
$=2.6\times10^8\text{m}$
View full question & answer→Question 753 Marks
The magnitude of gravitational field at distances $r_1$ and $r_2$ from the centre of a uniform sphere of radius $R$ and mass $M$ are $I_1$ and $I_2$ respectively. Find the ratio of $\left(\frac{I_1}{I_2}\right)$ if $r_1>R$ and $r_2<R$.
AnswerWhen $r_1 > R$, the point lies outside the sphere. Then sphere can be considered to be a point mass body whose whole mass can be supposed to be concentrated at its centre. Then gravitational intensity at a point distance $r_1$ from the centre of sphere will be, $\text{I}_1=\frac{\text{GM}}{\text{r}^2_1}\dots(\text{i})$ When $r_2 < R$, the point P lies inside the sphere. The unit mass body placed at P, will experience gravitational pull due to sphere of radius $r_2$_, whose mass is $\text{M}'=\frac{\text{M}\times\frac{4}{3}\pi\text{r}^3_2}{\frac{4}{3}\pi\text{R}^3}=\frac{\text{Mr}_2^3}{\text{R}^3}.$
Therefore the gravitational intensity at P will be $\text{I}_2=\frac{\text{GM r}_2^3}{\text{R}^3}\times\frac{1}{\text{r}_2^2}$
$=\frac{\text{GM r}_2}{\text{R}^3}\dots(\text{ii})$ So, $\frac{\text{I}_1}{\text{I}_2}=\frac{\text{GM}}{\text{r}^2_1}\times\frac{\text{R}^3}{\text{GM r}_2}=\frac{\text{R}^3}{\text{r}_1^2\text{r}_2}.$
View full question & answer→Question 763 Marks
Find an expression for the weight of a body at the centre of the Earth.
AnswerWe know that value of acceleration due to gravity at a depth'd below the surface of Earth is given by $\mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ At the centre of Earth $\mathrm{d}=\mathrm{R}$ and hence, $\mathrm{g}_{\text {centre }}=\mathrm{g}\left(1-\frac{\mathrm{R}}{\mathrm{R}}\right)=\mathrm{g}(1-1)=0 $
$\therefore$ Weight of a body at the centre of Earth $=\mathrm{mg}_{\text {centre }}=\mathrm{m} \times 0=0 \mathrm{It}$ means that at the centre of Earth a body will be weightless.
View full question & answer→Question 773 Marks
What is the angular velocity at any point on the equator so that the body feels weightlessness?
AnswerAt any point on the surface of earth, the acceleration due to gravity is, $\text{g}'=\text{g}\Big(1-\frac{\text{R}\omega^2}{\text{g}}\cos^2\alpha\Big),$ where $\alpha$ is the latitude. For equator, $\alpha=0$ To feel weightlessness $\text{g}'=0$ $\therefore\text{g}\Big(1-\frac{\text{R}\omega^2}{\text{g}}\Big)=0$ $\therefore\omega=\sqrt{\frac{\text{g}}{\text{R}}}$
View full question & answer→Question 783 Marks
A high jumper can jump $1.5m$ on earth. With the same effort, how high will he be able to jump on a planet whose density is one-third and radius is one-fourth of that of the earth?
AnswerLet $h_e$ be the height in metre, the man jumps on the earth and $h_p$ on the planet. If the effort is same, the P.E. gained is same. Therefore, $\text{Mg}_{\text{p}}\text{h}_{\text{p}}=\text{Mg}_{\text{e}}\text{h}_{\text{e}}$ $\text{h}_{\text{p}}=\frac{\text{g}_{\text{e}}\text{h}_{\text{e}}}{\text{g}_{\text{p}}}$ We know, $\text{g}_{\text{e}}=\frac{\text{GM}_{\text{e}}}{\text{R}_{\text{e}}}=\text{G}\frac{4}{3}\pi\text{R}_{\text{e}}\rho_{\text{e}}$ $\frac{\text{g}_{\text{e}}}{\text{g}_{p}}=\frac{\text{R}_{\text{e}}\rho_{\text{e}}}{\text{R}_{\text{e}}\rho_{\text{e}}}=12$ $\therefore\text{h}_{\text{p}}=12\times\text{h}_{\text{e}}=12\times1.5=18\text{m}$
View full question & answer→Question 793 Marks
A particle is projected vertically upwards from the surface of Earth of radius R with a kinetic energy equal to half of the minimum value needed for it to escape. Find the height to which it rises above the surface of Earth.
AnswerWe know that escape velocity from the surface of Earth is given by, $\text{v}_\text{es}=\sqrt{\frac{2\text{GM}}{\text{R}}}$ and corresponding K.E. of a body $\text{K}_\text{es}=\frac{1}{2}\text{mv}^2_\text{es}=\frac{\text{GMm}}{\text{R}}$ As in present problem, the body is projected from the surface of Earth with a kinetic energy half of that needed to escape from Earth's surface, hence Initial kinetic energy of body $\text{K}=\frac{\text{K}_\text{es}}{2}=\frac{\text{GMm}}{2\text{R}}$ and its potential energy at surface of Earth $\text{U}=-\frac{\text{GMm}}{\text{R}}$ $\therefore$ Total initial energy of body $\text{K}+\text{U}=\frac{\text{GMm}}{2\text{R}}-\frac{\text{GMm}}{\text{R}}$ $=-\frac{\text{GMm}}{2\text{R}}$ Let the body goes up to a maximum height h from surface of Earth, where its final K.E. = 0 and $\text{P.E.}=-\frac{\text{GMm}}{(\text{R}+\text{h})}$ $\therefore$ Total energy now $=0-\frac{\text{GMm}}{(\text{R}+\text{h})}=-\frac{\text{GMm}}{(\text{R}+\text{h})}$ From conservation law of mechanical energy, we have $-\frac{\text{GMm}}{2\text{R}}=-\frac{\text{GMm}}{(\text{R}+\text{h})}$ On simplification it leads to the result h = R.
View full question & answer→Question 803 Marks
If a satellite is revolving around a planet of mass M in an elliptical orbit of semi major axis a, show that orbital speed v of the satellite when at a distance r from the focus will be given by $\text{v}^2=\text{GM}\Big(\frac{2}{\text{r}}-\frac{1}{\text{a}}\Big)$
AnswerIn case of elliptical orbit of a satellite, its mechanical energy (i.e., sum of K.E. and P.E.) remains constant at any position of satellite in the orbit, which is given by K.E. + P.E. $=\frac{\text{GMm}}{2\text{a}}$ If at a position r, v is the orbital speed of satellite, then $\text{K.E.}=\frac{1}{2}\text{mv}^2$ $\text{and }\text{P.E.}=-\frac{\text{GMm}}{\text{r}}\dots(2)$ From (1) and (2), we have $\frac{1}{2}\text{mv}^2-\frac{\text{GMm}}{\text{r}}=-\frac{\text{GMm}}{2\text{a}}$ $\text{or }\text{v}^2=\text{GM}\Big[\frac{2}{\text{r}}-\frac{1}{\text{a}}\Big]$
View full question & answer→Question 813 Marks
Calculate the minimum energy required to launch a 250kg satellite from earth's surface at an altitude of 2R when R is the radius of the earth and is equal to 6400km.
AnswerThe total energy of a satellite of mass m in a circular orbit of radius r is where r is $\frac{1}{2}\text{mv}^2-\text{G}\frac{\text{Mm}}{\text{r}}$Where r is measured from the centre of the earth. Total mechanical energy in the orbit is $\text{E}=\text{G}\frac{\text{mM}}{2\text{r}}-\text{G}\frac{\text{Mm}}{\text{r}}$
$=-\text{G}\frac{\text{mM}}{2\text{r}}$
$\text{r}=2\text{R}+\text{R}=3\text{R}$
$\text{E}=-\text{G}\frac{\text{mM}}{6\text{R}}$
The potential energy on the surface of the earth $=-\text{G}\frac{\text{mM}}{\text{R}}$ Minimum energy required $=-\frac{1}{6}\text{G}\frac{\text{mM}}{\text{R}}-\Big(-\text{G}\frac{\text{Mm}}{\text{R}}\Big)$ $=\frac{5}{6}\text{G}\frac{\text{mM}}{\text{R}}$ $=\frac{5}{6}\text{mg R}=\frac{5}{6}\times250\times9.8\times6.4\times10^6\text{J}$ $=1.3\times10^{10}\text{J}.$
View full question & answer→Question 823 Marks
What is the period of revolution of Neptune around the sun, given that the diameter of its orbit is 30 times the diameter of the earth's orbit around the sun, both orbits being assumed to be circular? Taking the moon's orbit around earth to be r and mass of earth 81 times the mass of the moon, find the position of the point from the earth where the net gravitational field is zero.
AnswerAccording to Kepler's law, $\text{T}^2\propto\text{r}^3$ $\text{T}^2_2=\text{T}^2_1\Big(\frac{\text{r}^2}{\text{r}_1}\Big)^3;$Taking $\text{T}_1=1\text{ year}$
$\text{T}^2_2=(1)^2\ (30)^3=27000$ or $\text{T}_2=\sqrt{27000}=164.3\text{ year}.$
View full question & answer→Question 833 Marks
Two stationary particles of masses $M_1$ and $M_2$ are a distance d apart. A third particle lying on the line joining the particles, experiences no resultant gravitational force. What is the distance of this particle from $M_1$?
AnswerThe force on m towards $M_1$ is $\text{F}=\text{G}\frac{\text{M}_1\text{m}}{\text{r}^2}$ The force on m toward $M_2$ is $\text{F}=\text{F}\frac{\text{M}_2\text{m}}{(\text{d}-\text{r})^2}$
Equating two forces, we have, $\text{G}=\frac{\text{M}_1\text{m}}{\text{r}^2}=\text{G}\frac{\text{M}_2\text{m}}{(\text{d}-\text{r})^2}$ $\Big(\frac{\text{d}-\text{r}}{\text{r}}\Big)^2=\frac{\text{M}_2}{\text{M}_1}$ or $\frac{\text{d}}{\text{r}}-1=\frac{\sqrt{\text{M}_2}}{\sqrt{\text{M}_1}}$ $\Rightarrow\frac{\text{d}}{\text{r}}=\frac{\sqrt{\text{M}_2}+\sqrt{\text{M}_1}}{\sqrt{\text{M}_1}}$ So, distance of an particle from m is, $\text{r}=\text{d}\Big(\frac{\text{M}_1}{\sqrt{\text{M}_1}+\sqrt{\text{M}_2}}\Big)$ View full question & answer→Question 843 Marks
Find the angular velocity for an object to experience weightlessness at the equator of earth. Under this condition, also find the duration of a day.
AnswerDue to the rotation $\text{g}'=\text{g}\Big(1-\frac{\text{R}\omega^2}{\text{g}}\cos^2\theta\Big)$ where $\theta$ is the latitude. For weightlessness, g' = 0. At the equator, $\theta=0^\circ$ $\therefore1-\frac{\text{R}\omega^2}{\text{g}}\cos^20^\circ=0$ $\text{or }\omega=\sqrt{\frac{\text{g}}{\text{R}}}$ $\Rightarrow\omega=\sqrt{\frac{10}{6400\times10^3}}=\frac{1}{800}$ $\omega=\frac{2\pi}{\text{T}}$ $\Rightarrow=\frac{2\text{x}}{\omega}=\frac{2\times3.14}{\frac{1}{800}}$ $=5024\text{sec.}$
View full question & answer→Question 853 Marks
If the radius of the earth were increased by a factor of 3, by what factor would its density have to be changed to keep 'g' the same?
Answer$\text{g}=\frac{\text{GM}}{\text{R}^2}$ Let $\rho$ be the density of earth. $\rho=\frac{\text{M}}{\text{Volume of earth}}=\frac{\text{M}}{\frac{4}{3}\pi\text{R}^3}$ $\text{or }\text{M}=\frac{4}{3}\pi\text{R}\rho$ $\text{g}=\frac{\text{G}}{\text{R}^2}\times\frac{4}{3}\pi\text{R}\rho=\frac{4}{3}\pi\text{G}\rho\text{R}$ Since $\frac{4}{3},\pi,\text{G}$ are constants. For no change in value of $'\text{g}'\text{R}\propto\frac{1}{\rho}$ If R is made $3\text{R, }\rho$ must become $\frac{\rho}{3}.$
View full question & answer→Question 863 Marks
The planet Mars has two moons, phobos and Deimos.
- Phobos has a period $7$ hours, $39$ minutes and an orbital radius of $9.4 \times 10^3km$. Calculate the mass of mars.
- Assume that earth and mars move in circular orbits around the sun, with the Martian orbit being $1.52$ times the orbital radius of the earth. What is the length of the Martian year in days?
Answer
- The Sun's mass replaced by the martian mass $M_m$.
$\text{T}^2=\frac{4\pi^2}{\text{GM}_\text{m}}\text{R}^3$
$\text{M}_\text{m}=\frac{4\pi^2}{\text{G}}\times\frac{\text{R}^3}{\text{T}^2}$
$\text{M}_\text{m}=\frac{4\times(3.14)^2\times(9.4)^3\times10^{18}}{6.67\times10^{-11}\times(459\times60)^2}$
$=\frac{4\times(3.14)^2\times(9.4)^3\times10^{18}}{6.67\times(4.59\times6)^2\times10^{-5}}$
$=6.48\times10^{23}\text{kg}$
- Using Kepler's third law,
$\frac{\text{T}^2_\text{M}}{\text{T}^2_\text{E}}=\frac{\text{R}^3_\text{MS}}{\text{R}^3_\text{ES}}$
where $R_{MS} (R_{ES})$ is the Mars $($Earth$) -$ Sun distance.
$\text{T}_\text{M}=\Big(\frac{\text{R}_\text{MS}}{\text{R}_\text{ES}}\Big)^\frac{3}{2}\times\text{T}_\text{E}$
$=(1.52)^\frac{3}{2}\times365=684\text{ days}$ View full question & answer→Question 873 Marks
Two bodies of m and 4m are placed at a distance. The gravitational field is zero at a point on the line joining the two masses. What will be the gravitational potential at this point?
Answer
$\Rightarrow\text{3x}=\text{r}$ $\Rightarrow\text{x}=\frac{\text{r}}3{}$ $\therefore$ The gravitational potential $=\frac{-\text{Gm}}{\frac{\text{r}}{3}}\frac{-\text{(4m)}}{\frac{2\text{r}}{3}}$ $=\frac{-3\text{Gm}}{\text{r}}\frac{-6\text{Gm}}{\text{r}}=\frac{-9\text{Gm}}{\text{r}}$ View full question & answer→Question 883 Marks
What is escape velocity. Derive an expression for the same.
AnswerThe minimum velocity required to escape from the gravitational force of earth is called escape velocity. Total energy is the sum of P.E. and K.E. $\text{T.E}=-\frac{\text{GMm}}{\text{R}}+\frac{1}{2}\text{mv}^2$ To escape K.E should be greater than P.E., i.e.. $\frac{1}{2}\text{mv}^2\geq-\frac{\text{GMm}}{\text{R}}$ $\text{v}_{\text{e}}=\sqrt{2\frac{\text{GM}}{\text{R}}}=\sqrt{2\text{gR}}$
View full question & answer→Question 893 Marks
Draw areal velocity versus time graph for mars.
AnswerAreal velocity of Mars revolving around the Sun does not change with time according to Kepler’s law, i.e. it is constant with time. Then graph of a real velocity versus time is a straight line parallel to time axis.
View full question & answer→Question 903 Marks
If a body is taken to a height equal to the radius of earth from its surface, how much the weight of a body will decrease?
AnswerOn the surface of the earth,$\text{g}=\text{G}.\frac{\text{M}}{\text{R}^2}\dots(\text{i})$
and on the height where h = R $\text{g}'=\text{G}.\frac{\text{M}}{(\text{R}+\text{R})^2}=\text{G}.\frac{\text{M}}{4\text{R}^2}\dots(\text{ii})$ Dividing (ii) by (i) we get $\frac{\text{g}'}{\text{g}}=\frac{\text{G}.\frac{\text{M}}{4\text{R}^2}}{\text{G}.\frac{\text{M}}{\text{R}^2}}$ $\therefore\ \frac{\text{g}'}{\text{g}}=\frac{1}{4}\Rightarrow\ \text{g}'=\frac{\text{g}}{4}$ Hence the weight of the body will reduce to one-fourth of its original weight on the surface of the earth.
View full question & answer→Question 913 Marks
Can a satellite be in an orbit in a plane not passing through the earth's centre? Explain your answer.
Answer The centripetal force required for the orbital motion of the satellite is provided by the gravitational force of attraction. Gravitational force is a central force, i.e., it passes through the centre of mass of the earth and the satellite. Hence, the plane of orbit of the satellite has to pass through the earth’s centre.
View full question & answer→Question 923 Marks
A satellite is revolving around the earth, close to the surface of earth with a kinetic energy E. How much kinetic energy should be given to it so that it escapes from the surface of earth?
AnswerLet $v_0, v_e$ be the orbital and escape speeds of the satellite, then $\text{v}_\text{e}=\sqrt{2}\text{v}_0.$ Energy in the given orbit, $\text{E}_1=\frac{1}{2}\text{mv}_0^2=\text{E}\dots(\text{i})$ Energy for the escape speed, $\text{E}_2=\frac{1}{2}\text{mv}_\text{e}^2=\frac{1}{2}\text{m}(\sqrt{2}\text{v}_0)^2=2\text{E}$
$\therefore$ Energy required to be supplied =$E_2 - E_1 = E$.
View full question & answer→Question 933 Marks
A stationary object is released from a point P a distance 3R from the centre of the moon which has radius R and mass M. Calculate the speed of the object on hitting the moon.
AnswerGain in kinetic energy by the object = Loss in gravitational potential $\frac{1}{2}\text{mv}^2=\Big(-\frac{\text{GMm}}{\text{3R}}\Big)-\Big(-\frac{\text{GMm}}{\text{R}}\Big)$ $\Rightarrow\frac{1}{2}\text{v}^2=\frac{2\text{GM}}{3\text{R}}$ 
⇒ Speed of the object $=\text{v}=\sqrt{\frac{4\text{GM}}{3\text{R}}}$ View full question & answer→Question 943 Marks
Calculate the change in the energy of a $500kg$ satellite when it falls from an altitude of $200km$ to $199km$. If this change takes place during one orbit. Calculate the retarding force on the satellite. Given, mass of the earth = $6 \times 10^{24}kg$ and radius of the earth = $6400km$.
AnswerGiven, $M_e = 6 \times 10^{24}kg, r_e = 6400km r_1 = 6400 + 200 = 6600km = 6.6 \times 10^6m r_2 = 6400 + 199 = 6599km = 6.599 \times 10^6m$ Change in energy $=\text{GMm}\Big(\frac{1}{\text{r}_1}-\frac{1}{\text{r}_2}\Big)$
$=6.67\times10^{-11}\times6\times10^{24}\times500$
$\Big(\frac{1}{6.6\times10^6}-\frac{1}{6.599\times10^6}\Big)$
$=2\times10^{17}(1.5152\times10^{-7}-1.5154\times10^7)\text{J}$
$=-4\times10^6\text{J}$ If this occurs during one orbit, then the energy lost = force \times distance. If we take the distance as beiag the circumference of one orbit. Then, Retarding force $=\frac{4\times10^6}{2\pi\times6.6\times10^6}=\frac{4\times10^6}{2\times6.6\times3.14\times10^6}=0.1\text{N}$
View full question & answer→Question 953 Marks
A $400 kg$ satellite is in a circular orbit of radius $2 R_E$ about the Earth. How much energy is required to transfer it to a circular orbit of radius $4 R_E$ ? What are the changes in the kinetic and potential energies?
AnswerInitially,
$
E_i=-\frac{G M_E m}{4 R_E}
$
While finally
$
E_f=-\frac{G M_E m}{8 R_E}
$
The change in the total energy is
$
\begin{array}{c}
\Delta E=E_f-E_i \\
=\frac{G M_E m}{8 R_E}=\left(\frac{G M_E}{R_E^2}\right) \frac{m R_E}{8} \\
\Delta E=\frac{g m R_E}{8}=\frac{9.81 \times 400 \times 6.37 \times 10^6}{8}=3.13 \times 10^9 J
\end{array}
$
The kinetic energy is reduced and it mimics $\Delta E$, namely, $\Delta K=K_f-K_i=-3.13 \times 10^9 J$.
The change in potential energy is twice the change in the total energy, namely
$
\Delta V=V_f-V_i=-6.25 \times 10^9 J
$
View full question & answer→Question 963 Marks
Weighing the Earth : You are given the following data: $g=9.81\ ms ^{-2}, R_E=6.37 \times 10^6 m$, the distance to the moon $R =3.84 \times 10^8 m$ and the time period of the moon's revolution is $27.3$ days. Obtain the mass of the Earth $M_E$ in two different ways.
AnswerFrom Eq. $(7.12)$ we have
$M_E=\frac{g R_E^2}{G}$
$=\frac{9.81 \times\left(6.37 \times 10^6\right)^2}{6.67 \times 10^{-11}}$
$=5.97 \times 10^{24} \ kg .$
The moon is a satellite of the Earth. From the derivation of Kepler's third law $[$see Eq. $(7.38)]$
$T^2=\frac{4 \pi^2 R^3}{G M_E}$
$M_E=\frac{4 \pi^2 R^3}{G T^2}$
$=\frac{4 \times 3.14 \times 3.14 \times(3.84)^3 \times 10^{24}}{6.67 \times 10^{-11} \times(27.3 \times 24 \times 60 \times 60)^2}$
$=6.02 \times 10^{24} \ kg$
Both methods yield almost the same answer, the difference between them being less than $1 \%$.
View full question & answer→Question 973 Marks
The planet Mars has two moons, phobos and delmos.
$(i)$ phobos has a period $7$ hours$, 39$ minutes and an orbital radius of $9.4 \times 10^3 \ km$. Calculate the mass of mars.
$(ii)$ Assume that earth and mars move in circular orbits around the sun, with the martian orbit being $1.52$ times the orbital radius of the earth. What is the length of the martian year in days?
Answer$(i)$ We employ Eq. $(7.38)$ with the sun's mass replaced by the martian mass $M_m$
$T^2=\frac{4 \pi^2}{G M_m} R^3$
$M _m=\frac{4 \pi^2}{G} \frac{R^3}{T^2}$
$=\frac{4 \times(3.14)^2 \times(9.4)^3 \times 10^{18}}{6.67 \times 10^{-11} \times(459 \times 60)^2}$
$M _m=\frac{4 \times(3.14)^2 \times(9.4)^3 \times 10^{18}}{6.67 \times(4.59 \times 6)^2 \times 10^{-5}}$
$=6.48 \times 10^{23} \ kg .$
$(ii)$ Once again Kepler's third law comes to our aid,
$\frac{T_M^2}{T_E^2}=\frac{R_{M S}^3}{R_{E S}^3}$
where $R_{M S}$ is the mars $-$ sun distance and $R_{E S}$ is the earth$-$sun distance.
$\therefore T_M =(1.52)^{3 / 2} \times 365$
$ =684 \text { days }$
We note that the orbits of all planets except Mercury and Mars are very close to being circular.
For example, the ratio of the semiminor to semi-major axis for our Earth is, $b / a=0.99986$.
View full question & answer→Question 983 Marks
Find the potential energy of a system of four particles placed at the vertices of a square of side $l$. Also obtain the potential at the centre of the square.
AnswerConsider four masses each of mass $m$ at the corners of a square of side $l$; See Fig. $7.9$.
We have four mass pairs at distance $l$ and two diagonal pairs at distance $\sqrt{2} t$
Hence,
$W(r)=-4 \frac{G m^2}{l}-2 \frac{G m^2}{\sqrt{2} l}$
$=-\frac{2 G m^2}{l}\left(2+\frac{1}{\sqrt{2}}\right)=-5.41 \frac{G m^2}{l}$
The gravitational potential at the centre of the square $(r=\sqrt{2} l / 2)$ is
$
U(r)=-4 \sqrt{2} \frac{ Gm }{l} .
$ View full question & answer→