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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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Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to -

$R^{(\frac{n + 1}{2})}$

$R (\frac{n - 1}{2})$

$R^{n}$

$R (\frac{n - 2}{2})$

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Explanation

(a) ${mω}^{2} R \propto \frac{1}{R^{n}} \Rightarrow m (\frac{4 π^{2}}{T^{2}}) R \propto \frac{1}{R^{n}} \Rightarrow T^{2} \propto R^{n + 1} ∴ T \propto R^{(\frac{n + 1}{2})}$

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The orbital speed of an artificial satellite very close to the surface of the earth is $V_{o}$ . Then the orbital speed of another artificial satellite at a height equal to three times the radius of the earth is

$4 V_{0}$

$2 V_{0}$

$0.5 V_{0}$

$4 V_{0}$

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Explanation

If $r = R + h = R + 3 R = 4 R$ then $v = \frac{V_{o}}{2} = 0.5 V_{0}$

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If the radius of the earth were to shrink by 1% its mass remaining the same, the acceleration due to gravity on the earth's surface would -

Decrease by 2%

Remain unchanged

Increase by 2%

Increase by 1%

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Explanation

% increase in g = 2(% decrease in R) = 2x1% = 2%

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