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Physics 2 Marks Questions

Gravitation · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 86 questions.

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Question 12 Marks
Choose the correct alternative:
If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/ potential energy.
Answer
Kinetic energy.

Explanation:

Total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and potential energy (may be negative). At infinity, the gravitational potential energy of the satellite is zero. As the Earth-satellite system is a bound system, the total energy of the satellite is negative. Thus, the total energy of an orbiting satellite at infinity is equal to the negative of its kinetic energy.

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Question 22 Marks
Answer the following:
If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why?
Answer
Earth moon distance is very small as compared to earth-sun distance. Tidal effect is inversely proportional to the cube of the distance it means it is not governed by inverse square law like the gravitational force (which obeys inverse square law). Hence, tidal effect of moon is larger than that due to the sun.
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Question 32 Marks
Answer the following:
An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
Answer
Yes, Astronaut can hope to detect gravity if the size of th e spaceship is extremely large, then the magnitude of the gravity will become appreciable and hence the gravitational effect of the spaceship may been me measurable.
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Question 42 Marks
Answer the following:
You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
Answer
No. Electrical forces depend upon the nature of the intervening medium while the gravitational forces don’t depend upon the nature of the intervening medium. So, such shielding acts are not possible in case of gravitation i.e., gravity screens are not possible.
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Question 52 Marks
An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
Answer
In case the size of the space-station becomes large, the forces of gravity will become appreciable, and the astronaut can hope to detect it.
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Question 62 Marks
Earth's radius is about 6370km. A mass of 20kg is taken to a height of 160km above the earth's surface.
  1. What is the mass of the objects at that height?
  2. How much does the object weigh at this height?
Answer
  1. The mass of the object remain 20kg at that height.
  2. $\text{W}=\frac{\text{GMm}}{\text{r}^2}$

$\Rightarrow\frac{\text{W}_2}{\text{W}_1}=\frac{\text{r}^2_1}{\text{r}^2_2}$ since G, M and m are constant.

$\therefore\text{W}_2=9.8\times20\times\Big(\frac{6370}{6370+160}\Big)^2=186.5\text{N}$

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Question 72 Marks
What are the conditions under which a rocket fired from the earth, launches an artificial satellite of earth?
Answer
  1. Velocity acquired is more than the escape velocity to go out in space.
  2. Provide sufficient velocity to move in a path of its own.
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Question 92 Marks
If the choice of origin is shifted, what is the change in (i) Gravitational potential (ii) Potential energy?
Answer
  1. Remains dependent on the choice of origin.
  2. Potential energy is independent since it depends only on separation.
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Question 102 Marks
The asteroid pallas has an orbital period of 4.62 year. Find the semi-major axis of its orbit. Given, G = 6.67 × 10-11Nm2 kg-2,
Mass of the sun = 1.99 × 1030kg and 1year
= 3.156 × 107s
Answer
$\text{T}^2=\frac{4\pi^2\text{a}^3}{\text{GM}_\text{s}}$

$\text{a}=\Big[\frac{\text{GM}_\text{s}\text{T}^2}{4\pi^2}\Big]^\frac{1}{3}$

Substituting values, we get

a = 4.15 × 1011m

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Question 112 Marks
Two particles of equal mass 'm' go round a circle of radius R, under the action of their mutual gravitational attraction. What is the speed of each particle?
Answer

The gravitational force of attraction provides the necessary centripetal force.

$\frac{\text{mv}^2}{\text{R}}=\frac{\text{G}(\text{m})(\text{m})}{(2\text{R})^2}$

$\Rightarrow\text{v}^2=\frac{\text{Gm}}{4\text{R}}$

$\Rightarrow\text{v}=\frac{1}{2}\sqrt{\frac{\text{Gm}}{\text{R}}}$

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Question 122 Marks
A satellite does not need any fuel to circle around the earth. Why?
Answer
The gravitational force between satellite and the earth provides the centripetal force required by the satellite to move in a circular orbit. The satellite orbits around earth at such a higher height where air friction is neglible.
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Question 132 Marks
A planet reduces its radius by 1% with its mass remaining same. How acceleration due to gravity varies?
Answer
When mass is same, $\text{g}\propto\frac{1}{\text{R}^2}.$

$\therefore\frac{\Delta\text{g}}{\text{g}}=2\frac{\Delta\text{R}}{\text{R}}$

% variation of g is 2%.

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Question 142 Marks
If earth be at half of its present distance from the sun, how many days will there be in a year?
Answer
$\text{T}^2\propto\text{r}^3,\text{Since r}\rightarrow\frac{\text{r}}{2},\frac{\text{T}^2_1}{\text{T}^2}=\Big(\frac{1}{2}\Big)^3$
$\therefore\text{T}_1=\Big(\frac{1}{8}\Big)^{\frac{3}{2}}\text{T};\text{T}_1=\Big(\frac{1}{2\sqrt{2}}\Big)^3\text{T}$
$\text{i.e.}\text{T}_1=\frac{\text{T}}{16\sqrt{2}}$
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Question 152 Marks
Two masses each of 100kg are separated by 1 metre. Find the gravitational (i) field (ii) potential at the mid-point of the line joining them.
Answer
Gravitational field strength is the force experienced by unit mass kept at a point. Since it is a vector, we have to add vectorially.

$\therefore\text{E}_{\text{g0}}=\text{E}_{\text{gA}}+\text{E}_{\text{gB}}=0,$ Since the masses are equal and separation is same.

But gravitational potential $=-\frac{\text{Gm}}{\text{r}}$

Potential at O (mid-point) $=-\frac{\text{G}\times100}{0.5}-\frac{\text{G}\times100}{0.5}$

$=-\frac{200\text{G}}{0.5}=-400\text{G }\text{J/m}$

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Question 162 Marks
Give one example each of central force and non - central force.
Answer
Gravitational force, electrostatic force due to point mass and point charges are the examples of central force. Spin dependent nuclear force, magnetic force between two current carrying loops are the examples of non - central forces.
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Question 172 Marks
Under what circumstances would your weight become zero?
Answer
The weight will become zero under the following circumstances:
  1. During free fall.
  2. At the centre of the earth.
  3. In an artificial satellite.
  4. At a point where gravitational pull of earth is equal to the gravitational pull of the Moon.
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Question 182 Marks
A satellite is revolving just near the Earth's surface. Compute its orbital velocity. Given that radius of Earth R = 6400km and 3 = 9.8ms-2.
Answer
For a satellite revolving just near the Earth's surface, the orbital velocity has a magnitude given by,
$\text{v}_\text{orb}=\sqrt{\text{gR}}$
$\therefore\ \text{v}_\text{orb}=\sqrt{9.8\times6400\times1000}$
$=7.92\times10^3\text{ms}^{-1}$
$=7.92\text{km s}^{-1}$
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Question 192 Marks
The escape velocity v of a body depends upon:
  1. The acceleration due to gravity 'g' of the planet.
  2. The radius of the planet 'R'.
Establish dimensionally the relationship between them.
Answer
$\text{v}\propto\text{g}^{\text{a}}\text{R}^{\text{b}}$
$[\text{LT}_{-1}]=\text{K}[\text{LT}^{-2}]^\text{a}[\text{L}]^{\text{b}}$
$\text{a}=\frac{1}{2},\text{b}=\frac{1}{2}$
$\text{K}=\sqrt{2}$
$\therefore\text{v}=\sqrt{2\text{gR}}$
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Question 202 Marks
A spherical planet has mass Mp and diameter Dp. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to whom?
Answer
Force is given by,
$\text{F}=\frac{\text{GM}_\text{e}}{\text{R}^2}=\frac{\text{GM}_\text{P}\text{m}}{\Big(\frac{\text{D}_\text{P}}{2}\Big)}=\frac{4\text{GM}_\text{P}\text{m}}{\text{D}^2_\text{P}}$
$\frac{\text{F}}{\text{m}}=\frac{4\text{GM}_\text{P}}{\text{D}^2_\text{P}}$
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Question 212 Marks
Why does the earth have an atmosphere and the moon does not?
Answer
The escape velocity from the earth surface is about 11.2km/sec which is greater than the r.m.s. velocity of the air molecules. The escape velocity from the moon's surface is about 2.4km/sec which is less than the r.m.s. velocity of the air molecules. Hence, the earth has an atmosphere and the moon does not.
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Question 222 Marks
A 10kg mass is to be divided into two parts, such that the force of attraction between them is maximum. What is the mass of each portion?
Answer
Let m kg and (10 - m) kg be the mass of two parts separated by a distance r. The force is $\text{F}=\frac{\text{Gm}(10-\text{m})}{\text{r}^2}$
For force to be maximum, $\frac{\text{dF}}{\text{dm}}=0$
$\therefore\frac{\text{G}}{\text{r}^2}[\text{m}(-1)+(10-\text{m}\times1)=0]$
$\therefore\text{m}=10-\text{m}$
$\therefore\text{m}=5\text{kg}$
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Question 232 Marks
Determine the speed with which the earth would have to rotate on its axis so that a person on the equator would weigh 3/5th as much as at present. Take the equatorial radius as 6400km.
Answer
Acceleration due to gravity at the equator is,
$\text{g}_\text{e}=\text{g}-\text{R}\omega^2$
$\text{mg}_\text{e}=\text{mg}-\text{mR}\omega^2$
$\frac35\text{mg}=\text{mg}-\text{mR}\omega^2$ $\Big[\because\text{mg}_\text{e}=\frac35\text{mg}\Big]$
$\therefore\omega=\sqrt{\frac{2\text{g}}{5\text{R}}}=\sqrt{\frac{2\times9.8}{6\times6400\times10^3}}=7.8\times10^{-4}\text{rad/s}$
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Question 242 Marks
The radii of two planets are R and 2R respectively and their densities $\rho\text{ and }\frac{\rho}{2}$ respectively. What is the ratio of acceleration due to gravity at their surfaces?
Answer
$\text{g}=\frac{\text{GM}}{\text{R}^2}=\frac{\text{g}\frac{4}{3}\pi\text{R}^3\rho}{\text{R}^2}=\text{g}\frac{4}{3}\pi\text{R}\rho$
$\therefore\frac{\text{g}_1}{\text{g}_2}=\frac{\text{R}}{2\text{R}}.\frac{\rho}{\frac{\rho}{2}}=1$
$\therefore\text{g}_1:\text{g}_2=1:1$
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Question 252 Marks
The escape speed on the earth is 11.2km/ s.What is its value for a planet having double the radius and eight times the mass of the earth?
Answer
vp (escap speed on a planet) $=\sqrt{\frac{\text{GM}_\text{p}}{\text{R}_\text{p}}}$
ve (escape speed on the earth) $=\sqrt{\frac{\text{GM}_\text{e}}{\text{R}_\text{e}}}$
Clearly, $\frac{\text{v}_\text{p}}{\text{v}_\text{e}}=\sqrt{\frac{\text{M}_\text{p}}{\text{M}_\text{e}}\times\frac{\text{R}_\text{e}}{\text{R}}_\text{p}}=\sqrt{8\times\frac12}=2$
$\text{v}_\text{p}=2\text{v}_\text{e}22.4\text{km/s}$
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Question 262 Marks
The binding energy of a particle of mass m attached with the earth of mass M and radius R is $\frac{\text{GMm}}{\text{R}}.$ How much energy must be supplied to the particle so that it may escape the gravitational field of the earth? In what form this energy is supplied to the particle?
Answer
The particle will escape the gravitational field of the earth if the energy supplied to it is R equal to its binding energy.
Thus, energy supplied to the particle $=\frac{\text{GMm}}{\text{R}}.$ This energy is supplied in the form of the kinetic energy of the particle.
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Question 272 Marks
A planet moving along an elliptical orbit is closest to the Sun at a distance r1 and farthest away at a distance of r2. If v1 and v2 are the linear velocities at these points respectively, then find the ratio $\frac{\text{v}_1}{\text{v}_2}.$
Answer
From the law of conservation of angular momentum,
$\text{mr}_1\text{v}=\text{mr}_2\text{v}_2$
$\Rightarrow\text{r}_1\text{v}_1=\text{r}_2\text{v}_2$
$\frac{\text{v}_1}{\text{v}_2}=\frac{\text{r}_2}{\text{r}_2}$
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Question 282 Marks
What is a geostationary satellite? Is it same as geo synchronous satellite?
Answer
If the revolution period of a satellite is same as that of the earth, it is called a geo-stationary satellite. Yes, geo-synchronous satellites are geo-stationary satellites.
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Question 292 Marks
What is the acceleration due to gravity at the bottom of a sea 30km deep taking radius of the earth as 6.3 × 10km?
Answer
$\text{g}'=\text{g}\Big(1-\frac{\text{d}}{\text{R}}\Big)$
$=9.8\Big(1-\frac{30\times1000}{6.3\times(1000)^2}\Big) $
$=9.8\Big(1-\frac{1}{210}\Big)$
$=9.8\Big(\frac{209}{210}\Big)=9.75\text{ms}^{-2}$
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Question 302 Marks
Jf the kinetic energy of a satellite revolving around the earth in any orbit is doubled, then what will happen to it?
Answer
The total energy of a satellite in any orbit, E = -K, where
K is its KE in that orbit.
If its kinetic energy is doubled, i.e. an additional kinetic energy (K) is given to it, E = -K + K = 0 and the satellite will leave its orbit and go to infinity.
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Question 312 Marks
Determine the speed with which the earth has to rotate on its axis so that a person on the equator weigh $\Big(\frac{3}{5}\Big)^{\text{th}}$ as much as at present. Take the equatorial radius as 6400km.
Answer
$\text{Here, W = mg}$
$\text{and }\text{W}'=\Big(\frac{3}{5}\Big)\text{mg}$
$\text{As W}'=\text{W}-\text{mR}\omega^2$
$\text{So},\frac{3}{5}\text{mg}=\text{mg}-\text{m}\text{R}\omega^2$
$\text{or }\text{mR}\omega^2=\text{mg}-\frac{3}{5}\text{mg}=\frac{2}{5}\text{mg}$
$\text{or }\omega=\sqrt{\frac{2\text{g}}{5\text{R}}}=\sqrt{\frac{2\times9.8}{5\times6400\times10^3}}$
$=4.315\times10^{-4}\text{rad/s}.$
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Question 322 Marks
Calculate the gravitational force which Sun exerts on Jupiter.
Answer
Given: G = 6.67 × 10-11Nm2/ kg2
MJ = 1.9 × 1027kg
Ms = 1.99 × 1030kg
Mean distance = 7.8 × 10+11m
$\text{F}_\text{g}=\frac{\text{GM}_{\text{S}}\text{M}_\text{J}}{\text{r}^2}$
$=\frac{6.67\times10^{-11}\times1.99\times10^{30}\times1.9\times10^{27}}{(7.8\times10^{11})^2}$
$=4.145\times10^{23}\text{N}$
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Question 332 Marks
Is it possible to place a satellite, so that it is always over New Delhi? Why?
Answer
No, generally the geo-stationary satellites will be in the equatorial plane. Since New Delhi does not lie in this plane, it is not possible.
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Question 342 Marks
If suddenly the gravitational force of attraction between the earth and a satellite revolving around it becomes zero, what will happen to the satellite?
Answer
Satellite will move tangentially with constant speed equal to its orbital velocity at the time the force ceases to act.
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Question 352 Marks
How can you find the mass of earth, starting from the law of gravitation?
Answer
From law of gravitation, $\text{F} =\frac{\text{GMm}}{\text{r}^2}$If m is a mass on the surface of earth of radius R. The acceleration of the mass m on the surface of earth is g.
$\therefore$ Force experienced F = mg
Using the two forces we have, $\text{M}=\frac{\text{gR}^2}{\text{G}}$
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Question 362 Marks
What is the gravitational potential energy of a body at heighth from the earth surface?
Answer
Gravitational potential energy, i.e.,
$\text{U}_\text{h}=-\frac{\text{GMm}}{\text{R}+\text{h}}=-\frac{\text{gR}^2\text{m}}{\text{R}+\text{h}}$ $\Big[\text{where},\text{g}=\frac{\text{GM}}{\text{R}^2}\Big]$
$=-\frac{\text{gR}^2\text{m}}{\text{R}\Big(1+\frac{\text{h}}{\text{R}}\Big)}=-\frac{\text{mgR}}{1+\frac{\text{h}}{\text{R}}}$
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Question 372 Marks
The acceleration due to gravity on a planet is 1.96m s-2. If it is safe to jump from a height of 2m on the earth, then what will be the corresponding safe height on the planet?
Answer
The safety of a person depends upon the momentum with which the person hits the planet. Since, the mass of the person is constant, therefore the maximum velocity v is the limiting factor.
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Question 382 Marks
Define Escape velocity and find its value.
Answer
The velocity with which a body is escaped vertically upwards and it just cross the gravitational field is called escape velocity of the body. Its value is,
$\text{V}_\text{e}=\sqrt{2\text{gR}_\text{e}}=11.2\text{km/ s}$
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Question 392 Marks
An astronaut, by mistake, drops his food packet from an artificial satellite orbiting around the earth. Will it reach the surface of the earth? Why?
Answer
The food packet will not fall on the earth. As the satellite as well as astronaut were in a state of weightlessness, hence, the food packet, when dropped by mistake, will also start moving with the same velocity as that of satellite and will continue to move along with the satellite in the same orbit.
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Question 402 Marks
A mass M1 revolves around another mass M2 in a path of radius r, what is the angular momentum associated with M1?
Answer
Angular momentum L is the product of twice the mass and aerial velocity.
$\therefore\text{L}=2\text{M}\frac{\text{dA}}{\text{dt}}=2\text{M}_1\times\frac{\pi\text{r}^2}{\text{T}}$
$=2\text{M}_1\frac{\pi\text{r}^2\text{v}}{2\pi\text{r}}$
$=\text{M}_1\text{r}\sqrt{\frac{\text{GM}_2}{\text{r}}}$
$=\text{M}_1\sqrt{\text{GM}_2\text{r}}$
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Question 412 Marks
What is the height at which the value of g is the same as at a depth of $\frac{\text{R}}{2}?$
Answer
At depth $\frac{\text{R}}{2},\text{g}'=\text{g}\Big(1-\frac{\text{R}}{2\text{R}}\Big)=\frac{\text{g}}{2}$

At height $\text{x},\text{g}'=\text{g}\Big(1-\frac{2\text{x}}{\text{R}}\Big)$

$\therefore\text{g}\Big(1-\frac{2\text{x}}{\text{R}}\Big)=\frac{\text{g}}{2}$

$\frac{1}{2}=\frac{2\text{x}}{\text{R}}$

$\therefore\text{x}=\frac{\text{R}}{4}$

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Question 422 Marks
A body has a sense of weightlessness in a satellite revolving around the earth, why?
Answer
The astronauts and the satellite require the centripetal force to revolve around the earth. Their weight is used up in providing the necessary centripetal force. Hence an astronaut feels weightlessness in the space.
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Question 432 Marks
State Newton's law of gravitation in vector form.
Answer
The force of attraction between a pair of masses m1 and m2 separated by a length 'r' is given by.
$\overrightarrow{\text{F}_{12}}=-\frac{\text{Gm}_1\text{m}_2}{\text{r}_2}\hat{\text{r}}_{21}$
$[\text{F}_{12}\rightarrow\text{force on }1\text{ due to }2].$
$\hat{\text{r}}\rightarrow\text{Points from }2\text{ to }1.$
-ve sign shows that the force is attractive.
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Question 442 Marks
The moon takes about 27 days to complete one orbit around the earth. The orbit is nearly a circle of radius 3.8 × 108m. Calculate the mass of the earth from this data.
Answer
Let M and m be the mass of the earth and the moon respectively. The gravitational force of attraction provides the centripetal force.
$\frac{\text{GMm}}{\text{r}^2}=\frac{\text{mv}^2}{\text{r}}$
$\Rightarrow\text{M}=\frac{\text{v}^2\text{r}}{\text{G}}=\frac{\omega^2\text{r}^3}{\text{G}}$
$=\Big(\frac{2\pi}{27\times24\times3600}\Big)^2\times\frac{(3.8\times10^8)^3}{6.67\times10^{-11}}$
$=5.968\times10^{24}\text{kg}$
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Question 452 Marks
A rocket is fired with a velocity 0.6 times the escape velocity on the surface of earth. How high will it go from the surface?
Answer
Velocity provided = 0.6ve,
$\text{v}_{\text{e}}=\sqrt{\frac{2\text{GM}}{\text{R}}}$
Applying conservation,
$-\frac{\text{GMm}}{\text{R}}+\frac{1}{2}\text{m}(6\text{v}_{\text{e}})^2=-\frac{\text{GMm}}{\text{R}+\text{h}}$
Solve for h to get
$\text{h}=\frac{72}{28}\text{R}.$
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Question 462 Marks
Answer the following:
An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
Answer
Yes, Astronaut can hope to detect gravity if the size of th e spaceship is extremely large, then the magnitude of the gravity will become appreciable and hence the gravitational effect of the spaceship may been me measurable.
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Question 472 Marks
If a planet existed whose mass and radius were both half of those of the earth, what would be the value of the acceleration due to gravity on its surface as compared to what it is on the earth's surface?
Answer
We know,
$\text{g}=\frac{\text{GM}}{\text{R}^2}\dots(1)$
$\text{g}'=\frac{\text{GM}'}{(\text{R}')^2}\Big[\text{M}'=\frac{1}{2}\text{M,}\text{R}'=\frac{1}{2}\text{R}\Big]\dots(2)$
Form (1) and (2), we have
$\frac{\text{g}'}{\text{g}}=\frac{1}{2}2^2=2$
$\therefore\text{g}'=2\text{g}$
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Question 482 Marks
A black hole is a body from whose surface nothing may ever escape. What is the condition for a uniform spherical mass M to be a black hole? What should be the radius of such a black hole if its mass is the same as that of the Earth?
Answer
For a body to be a black hole, the escape velocity should be such that even light cannot escape. The limiting case for escape velocity is,
$\sqrt{\frac{2\text{GM}}{\text{R}}}\leq\text{c}(\text{speed of light})$
For our earth, M = Me = 6 × 1024kg,
$\text{R}=\frac{2\text{GM}}{\text{c}^2}=9\times10^{-2}\text{m or }9\text{cm}$
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Question 492 Marks
Does the concentration of the earth's mass near its centre change the variation of g with height compared with a homogeneous sphere, how?
Answer
Any change in the distribution of the earth's mass will not affect the variation of acceleration due to gravity with height. This is because for a point outside the earth, the whole mass of the earth is effective and the earth behaves as a homogeneous sphere.
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Question 502 Marks
Why are space crafts usually launched from west to east? Why is it more advantageous to launch rockets in the equatorial plane?
Answer
Earth rotates on its axis from west to east. A satellite launched from west to east will have the advantage of the additional velocity of the earth's rotation. The effect is maximum at the equator, hence it is most advantageous to launch the satellite from west to east on the equatorial plane.
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2 Marks Questions - Physics STD 11 Science Questions - Vidyadip