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A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass=5.98X 10²⁴ kg) have to be compressed to be a black hole?

10^-9m

10^-6 m

10^-2m

100 m

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Explanation

Problem Solving Strategy For the black hole, the escape speed is more than c (speed of light). We should compare the escape speed with the c (Note that the escape speed should be at least just greater than c)

$v_{e} = \sqrt{2 G M / R}' [R' - > N e w r a d i u s o f t h e e a r t h] c = \sqrt{2 G M / R'} [V_{e} \approx c]$

=> c²=2GM/R'

R'=2GM/c²= $\frac{2 x 6.67 x 10^{- 11} x 6 x 10^{24}}{9 \times 10^{16}}$ =8.89x10^-3=0.889x10^-2=10^-2 m

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Infinite number of bodies, each of mass 2 kg are situated on x-axis at distance 1m,2m,4m,8m, respectively from the origin. The resulting gravitational potential due to this system at the origin will be

-G

-8/3G

-4/3G

-4G

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Explanation

(d)The resulting gravitational potential,

$V = - 2 G [1 / 1 + 1 / 2 + 1 / 4 + 1 / 8 + . . .] V = - 2 G [1 + 1 / 2 + 1 / 2^{2} + 1 / 2^{3} . . .]$

V=-2G(1-1/2)^-1

V=-2G/(1-1/2)

=-2G/(1/2)

=-4G

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The height at which the weight of a body becomes 1/16th, its weight on the surface of earth(radius R), is

5 R

15 R

3 R

4 R

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Explanation

According to question $\frac{G M m}{{(R + h)}^{2}} = \frac{1}{16} \frac{G M m}{R^{2}}$

$\frac{1}{{(R + h)}^{2}} = \frac{1}{16 R^{2}}$

$o r \frac{R}{R + h} = \frac{1}{4}$

$\frac{R + h}{R} = 4$

h=3R

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A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth.The time period of another satellite in hours at a height of 2R from the surface of the earth is

6 $\sqrt{2}$

6/ $\sqrt{2}$

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Explanation

From Keplar third's law

$T^{2} \propto r^{3}$

Hence, $T_{1}^{2} \propto r_{1}^{3}$

and $T_{2}^{2} \propto r_{2}^{3}$

So, $\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{r_{2}^{3}}{r_{1}^{3}}$

$= \frac{{(3 R)}^{3}}{{(6 R)}^{3}}$

or $\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{1}{8}$

$T_{2}^{2} = \frac{1}{8} T_{1}^{2}$

$T_{2} = \frac{24}{2 \sqrt{2}} = 6 \sqrt{2} h$

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If $v_{e}$ is escape velocity and $v_{0}$ is orbital

velocity of a satellite for orbit close to the

Earth's surface, then these are related by

$v_{o} = \sqrt{2 v_{e}}$

$v_{o} = v_{e}$

$v_{e} = \sqrt{2 v_{o}}$

$v_{e} = \sqrt{2} v_{o}$

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Explanation

Orbital velocity of satellite

$v_{o} = \sqrt{\frac{GM}{R}}$ ....(i)

and escape velocity of satellite

$v_{e} = \sqrt{\frac{2 GM}{R}}$ ...(ii)

Hence, $v_{e} = \sqrt{2} v_{o}$

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A planet moving along an elliptical orbit is closest to the sun at a distance $r_{1}$ and farthest away at a distance of $r_{2}$ . If $v_{1} and v_{2}$ are the linear velocities at these points respectively, then the ratio $\frac{v_{1}}{v_{2}}$ is

$r_{2} / r_{1}$

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

$r_{1} / r_{2}$

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

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Explanation

From the law of conservation of angular momentum

${mr}_{1} v_{1} = {mr}_{2} v_{2} r_{1} v_{1} = r_{2} v_{2} \frac{v_{1}}{v_{2}} = \frac{r_{2}}{r_{1}}$

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A body projected vertically from the earth reaches a high equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatest

at the instant just before the body hits the earth

it remains constant all through

at the instant just after the body is projected

at the highest position of the body

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Explanation

We know

$p = F \cdot v = F v \cos θ$

So just before hitting $θ$ is zero and both F and $v$ are maximum.

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A particle of mass m is thrown upwards from the surface of the earth with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth is

$\sqrt{\frac{2 G M}{R}}$

$\sqrt{\frac{2 G M}{R^{2}}}$

$\sqrt{2 g R^{2}}$

$\sqrt{\frac{2 G M}{R^{2}}}$

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Explanation

Escape velocity

$V_{e} = \sqrt{2 g R} V_{e} = \sqrt{\frac{2 G M}{R^{2}} R} (∵ G M = g R^{2}) V_{e} = \sqrt{\frac{2 G M}{R}}$

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The radii of circular orbits of two satellites A and B of the earth are 4R and R, respectively.If the speed of satellite A is 3v, then speed of satellite B will be

3v/4

12v

3v/2

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Explanation

Orbital velocity of satellite v= $\sqrt{\frac{G M}{r}}$

$\Rightarrow \frac{v_{A}}{v_{B}} = \sqrt{\frac{r_{B}}{r_{A}}} = \sqrt{\frac{R}{4 R}} = \frac{1}{2}$

$∴ \frac{v_{A}}{v_{B}} = \frac{3 v}{v_{B}} = \frac{1}{2}$

$∴ v_{B} = 6 v$

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The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius $R_{1}$ to another of radius $R_{2} (R_{2} > R_{1})$ is

$G m M (\frac{1}{R_{1}^{2}} - \frac{1}{R_{2}^{2}})$

$G m M (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$2 G m M (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{1}{2} G m M (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

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Explanation

$Total KE required to change the orbit will be equal to change in total energy as PE as well as KE, both are changing . ∆ E = (- \frac{G M m}{2 R_{2}}) - (- \frac{G M m}{2 R_{1}}) {KE}_{extra} = \frac{G M m}{2} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

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