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Gravitation MCQs for NEET — Physics Questions with Answers

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Two planets move around the sun. The periodic times and the mean radii of the orbits are $T_{1}, T_{2}$ and $r_{1}, r_{2}$ respectively. The ratio $T_{1} / T_{2}$ is equal to

${(r_{1} / r_{2})}^{1 / 2}$

$r_{1} / r_{2}$

${(r_{1} / r_{2})}^{2}$

${(r_{1} / r_{2})}^{3 / 2}$

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Explanation

(d) $T^{2} \propto r^{3} \Rightarrow \frac{T_{1}}{T_{2}} = {(\frac{r_{1}}{r_{2}})}^{3 / 2}$

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Orbital velocity of earth's satellite near the surface is 7 km/s. When the radius of the orbit is 4 times than that of earth's radius, then orbital velocity in that orbit is

3.5 km/s

7 km/s

72 km/s

14 km/s

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Explanation

(a) $v \propto \frac{1}{\sqrt{r}} .$ If orbital radius becomes 4 times then orbital velocity will

become half. i.e. $\frac{7}{2} = 3.5 k m / s$

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The largest and the shortest distance of the earth from the sun are $r_{1}$ and $r_{2}$ , its distance from the sun when it is at the perpendicular to the major axis of the orbit drawn from the sun

$\frac{r_{1} + r_{2}}{4}$

$\frac{r_{1} r_{2}}{r_{1} + r_{2}}$

$\frac{2 r_{1} r_{2}}{r_{1} + r_{2}}$

$\frac{r_{1} + r_{2}}{3}$

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Explanation

$r_{1} = (1 + e) a a n d r_{2} = (1 - e) a \Rightarrow a = \frac{r_{1} + r_{2}}{2} a n d r_{1} r_{2} = (1 - e^{2}) a^{2}$

where a = semi major axis

b = semi minor axis

e = eccentricity

Now required distance = semi latus rectum= $\frac{b^{2}}{a}$

= $\frac{a^{2} (1 - e^{2})}{a} = \frac{(r_{1} r_{2})}{(r_{1} + r_{2}) / 2} = \frac{2 r_{1} r_{2}}{r_{1} + r_{2}}$

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Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between planet and star is proportional to $R^{\frac{- 5}{2}}$ , then $T^{2}$ is proportional to

$R^{3}$

$R^{7 / 2_{}^{}}$

$R^{5 / 2}$

$R^{3 / 2}$

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Explanation

(b) For revolution of planet centripetal force is provided by gravitational force of attraction ${mω}^{2} R \propto R^{- 5 / 2} \Rightarrow \frac{1}{T^{2}} \propto R^{- 7 / 2} \Rightarrow T^{2} \propto R^{7 / 2}$

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A mass M is split into two parts, m and (M–m), which are then separated by a certain distance. What ratio of m/M maximizes the gravitational force between the two parts

1/3

1/2

1/4

1/5

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Explanation

(b) $F = \frac{Gm (M - m)}{r^{2}}$

For maximum force $\frac{dF}{dm} = 0$

$\Rightarrow \frac{d}{dm} (\frac{GmM}{r^{2}} - \frac{{Gm}^{2}}{r^{2}}) = 0 \Rightarrow M - 2 m = 0 \Rightarrow \frac{m}{M} = \frac{1}{2}$

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Two identical satellites are at R and 7R away from earth surface, the wrong statement is (R = Radius of earth)

Ratio of total energy will be 4

Ratio of kinetic energies will be 4

Ratio of potential energies will be 4

Ratio of total energy will be 4 but ratio of potential and kinetic energies will be 2

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Explanation

(d) Orbital radius of satellites $r_{1} = R + R = 2 R$

$r_{2} = R + 7 R = 8 R$

$U_{1} = \frac{- G M m}{r_{1}}$ and $U_{2} = \frac{- G M m}{r_{2}}$
   $K_{1} = \frac{G M m}{2 r_{1}}$ and $K_{2} = \frac{G M m}{2 r_{2}}$
$E_{1} = \frac{G M m}{2 r_{1}}$ and $E_{2} = \frac{G M m}{2 r_{2}}$
   $∴ \frac{U_{1}}{U_{2}} = \frac{K_{1}}{K_{2}} = \frac{E_{1}}{E_{2}} = 4$

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For a satellite escape velocity is 11 km/s. If the satellite is launched at an angle of 60° with the vertical, then escape velocity will be

11 km/s

$11 \sqrt{3} k m / s$

$\frac{11}{\sqrt{3}} k m / s$

33 km/s

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Explanation

(a) Escape velocity is same for all angles of projection.

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The mean radius of the earth is R, its angular speed on its own axis is $ω$ and the acceleration due to gravity at earth's surface is g. The cube of the radius of the orbit of a geostationary satellite will be -

$R^{2} g / ω$

$R^{2} ω^{2} / g$

$R g / ω^{2}$

$R^{2} g / ω^{2}$

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Explanation

(d) Orbital velocity $v_{o} = \sqrt{\frac{G M}{r}} = \sqrt{\frac{g R^{2}}{r}}$ and $v_{o} = r ω$

This gives $r^{3} = \frac{R^{2} g}{ω^{2}}$

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A satellite whose mass is M, is revolving in circular orbit of radius r around the earth. Time of revolution of satellite is

$T \propto \frac{r^{5}}{G M}$

$T \propto \sqrt{\frac{r^{3}}{G M}}$

$T \propto \sqrt{\frac{r}{G M^{2} / 3}}$

$T \propto \sqrt{\frac{r^{3}}{G M^{2} / 4}}$

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Explanation

(b) $T = \frac{2 πr}{v} . . . (1) a n d \frac{m v^{2}}{r} = \frac{G m_{e} M}{r^{2}} \Rightarrow v^{2} = \frac{G m_{e}}{r} o r, v = \sqrt{\frac{G m_{e}}{r}} . . . (2)$

Putting the value of v from equation (2) in equation (1)

$T = \frac{2 πr}{\sqrt{\frac{G m_{e}}{r}}} = 2 π \sqrt{\frac{r^{3}}{{Gm}_{e}}} \Rightarrow T α \sqrt{\frac{r^{3}}{{GM}_{e}}}$

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The periodic time of a communication satellite is

6 hours

12 hours

18 hours

24 hours

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Explanation

(d) Communication satellite is a kind of geostationary satellite so its time period is 24 hours.

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