Physics M.C.Q (1 Marks)

1. Planets will not have elliptic orbits.
2. Circular orbits of planets is not possible.
3. Projectile motion of a stone thrown by hand on the surface of the earth will be approximately parabolic.

Explanation:

If the law of gravitation becomes an inverse cube law instead of inverse square law, then for a planet of mass m revolving around the sun of mass M, we can write

$\text{F}=\frac{\text{GMm}}{\text{R}^3}=\frac{\text{mv}^2}{\text{R}}$ (where R is the radius of orbiting planet)

$\Rightarrow\ \text{Orbital speed}=\frac{\sqrt{\text{GM}}}{\text{R}}\Rightarrow\text{V}\propto\frac{1}{\text{ R}}$

Time period of revolution of a planet

$\text{T}=\frac{2\pi\text{R}}{\text{v}}=\frac{2\pi\text{R}}{\frac{\sqrt{\text{GM}}}{\text{R}}}=\frac{2\pi\text{R}^2}{\sqrt{\text{GM}}}$

$\Rightarrow\ \text{T}^2\propto\text{R}^4$

Hence, orbit will not be elliptical.

$[$For elliptical orbit $\text{T}^2\propto\text{R}^3]$

The circular orbits of the planets is not possible according to new law of gravitation.

As force $\text{F}=\Big(\frac{\text{GM}}{\text{R}^3}\Big)\text{m}$ = g'm

where, g' $=\frac{\text{GM}}{\text{R}^3}$

As g', acceleration due to gravity is constant, hence path followed by a projectile will be approximately parabolic, $(\text{as T}\propto\text{R}^2).$

Also, gravitational force inside a spherical shell of uniform density will have some value. So, only option (d) is incorrect.

1. Acceleration due to gravity decreases with increasing altitude.

3. Acceleration due to gravity increases with increasing latitude.

Explanation:

1. The acceleration due to gravity at an altitude (height), gh $=g(1-2).$

Increasing height (h) decreases the value of gh.option (a) is correct.

2. Assuming the earth to be a sphere of uniform density, the acceleration due to gravity at a particular depth (d), $\text{gd}=\text{g}\Big(1-\frac{\text{d}}{\text{R}}\Big).$

Increasing depth (d) decreases the value of ga . Option (b) is incorrect.

3. If $\lambda$ is latitude on earth then $\text{g}\lambda=\text{g}-\omega^2\text{R}\cos^2\lambda$

As value of cos decrease from 0°to 90° (from 1 to 0). The acceleration due to gravity increases from equator $(\lambda=0^0)$ to pole $(\lambda=90^0).$ Option (c) is correct.

4. The acceleration due to gravity on surface of earth is $\text{g}=\frac{\text{GM}_\in}{\text{R}^2_\in}$ So g on earth depends on mass of earth. Option (d) is incorrect.

3. M will move towards 2M.

Explanation:

Let FBC be the force experienced by mass m at point B due to mass M at point C, and FBA be the force experienced by mass m at point B due to mass 2M at point A.

$\text{F}_{\text{BC}}=\text{G}\frac{\text{mM}}{(\text{BC})^2}...(1)$

$\text{F}_\text{AB}=\text{G}\frac{\text{m2M}}{(\text{BA})^2}...(2)$

Suppose AB = x, then $\text{x}=\frac{1}{2}\text{ (BC)}\text{or}\text{(BC)}=2\text{x}$

Substituting the values of AB and BC in the above equations (1) and (2)

$\text{F}_{\text{BC}} =\text{G}\frac{\text{mM}}{(\text{2x})^2}$

$\text{F}_\text{BA}=\text{G}\frac{\text{m2M}}{(\text{x})^2}$

As $\text{F}_{\text{BA}}>\text{F}_{\text{BC}},$ m will move towards point A (position of particle with mass 2M).

1. Walking on ground would became more difficult.

3. Raindrops will fall much faster.
4. Airplanes will have to travel much faster.

Explanation:

If the gravitational constant G becomes 10 times larger in magnitude.

$\text{G}'=10\text{G}$

Gravitational field due to the earth

$\text{g}'=\frac{\text{G}'\text{M}_\text{e}}{\text{r}^2}$

$=\frac{10\text{GM}_\text{e}}{\text{r}^2}=10\text{g}$

Weight of a person $=\text{mg}'=\text{m}\times10\text{g}=10\text{mg}$

Force on the man due to sun, $\text{F}=\frac{\text{GM}'_\text{s}\text{m}}{\text{r}^2}$

Mass of the sun $\text{M}'_\text{s}=\frac{1}{10}\text{M}_\text{s}\Rightarrow10\text{M}'_\text{s}=\text{M}_\text{s}$

$\text{F}=\frac{\text{GM}_\text{s}\text{m}}{\text{10r}^2}$

Weight of person becomes 10 times larger so it will be more difficult to walk. Option (a) is correct.

As g’ = 10g, the acceleration due to gravity changes. Option (b) is incorrect.

The terminal velocity $\text{v}_\text{T}\propto\text{g}$ and g', g' = 10g, the terminal velocity increases 10 times.

Hence the rain drops falls 10 times faster. Option (c) is correct.

As the g’ = 10g, to overcome the increase gravitational force and in order to maintain the speed the aeroplane will have to travel much faster. Option (d) is correct.

1. The acceleration due to gravity on earth will be different for different objects.

3. Only the third law will become invalid.
4. For n negative, an object lighter than water will sink in water.

Explanition:

According to the problem,

$\text{F}_1=-\text{F}_2=-\frac{\text{r}^{12}}{\text{r}^3_{12}}\text{GM}_0^2\Big(\frac{\text{m}_1\text{m}_2}{\text{M}^2_0}\Big)^\text{n}$

$\Rightarrow\ \vec{\text{r}}_{12}=\text{r}_1-\text{r}_2$

$(\text{a})\text{F}=\frac{\text{GM}^{2(1-\text{n})}_0(\text{M})^\text{n}}{\text{r}^2_{12}}(\text{m}_1\text{m}_2)^\text{n}$

Take, m₁ = M (mass of earth), m₂ = m (mas of the object), r₁₂ R (radius of earth)

Therefore, $\text{F}=\Big(\frac{\text{GM}_0^{(2-2\text{n})}(\text{M})^\text{n}}{\text{R}^2}\Big)\text{m}^\text{n}=\text{Km}^\text{n}$

Where K is the constant or the term in the bracket is regarded as constant.

As $\text{F}=\text{mg},\text{so g}=\text{K m}^\text{n-1}$, hence g depends upon the mass of object.

Since, g depends upon position vector and mass of object, hence it will be different for different objects. As g is not constant, hence constant of proportionality will not be constant in Kepler’s third law.

Hence, Kepler’s third law will not be valid.

As the force is of central nature,

$\Big[\because\text{Force}\propto\frac{1}{\text{r}^2}\Big]$

Hence, first two Kepler’s laws will be valid. Hence option (b) in incorrect and (c) is correct.

(d) When n is negative, F = K / Mⁿ Or we can say that F is inversely proportional to mass. This implies that lighter bodies will experience a greater force than the heavier bodies and vice versa. Hence, object lighter than water will sink in water.

3. We will be going around but not strictly in closed orbits.
4. After sufficiently long time we will leave the solar system.

Explanition:

We know that gravitational force exists between the earth and the sun.

$\text{F}_\text{G}=\frac{\text{G}(\text{M}_\text{S}\times\text{m}_\text{e})}{\text{r}^2}$ Where M_S is mass of the sun and m_e is mass of the earth.

This provides the necessary centripetal force for the circular orbit of the earth around the sun. As G decreases with time, the gravitational force F_G will become weaker with time. As F_G is changing with time due to it, the earth will be going around the sun not strictly in closed orbit and radius also increases, since the attraction force is getting weaker.
Hence, after long time the earth will leave the solar system.

1. The torque is zero.

Explanation:

As the earth is revolving around the sun in a circular motion (approximately in actual the path of earth around the sun is elliptical) due to gravitational attraction. When we consider the earth-sun as a single system and we are taking earth as a sphere of uniform density. Then the gravitational force (F) will be of radial nature, i.e. angle between position vector r and force F is zero. So, torque

$|\vec{\tau}|=|\vec{\text{r}}\times\vec{\text{F}}|=\text{r F}\sin0^0=0.$

3. Of viscous forces causing the speed of satellite and hence height to gradually decrease.

Explanation:

The P.E. of satellite orbiting in orbit of radius r due to earth of mass M is $\Big(\frac{-\text{GM}}{2\text{r}}\Big)$ negative sign shows the force of attraction between satellite and earth. Energy (P.E.) is continuously reduced due to atmospheric friction, the radius of the orbit or height decreases gradually, and ultimately it comes back to earth with increasing speed and burns in the atmosphere.