1

### JEE Main 2019 (Online) 9th January Evening Slot

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 $\times$ 103 km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is
A
1.6 $\times$ 103 km
B
3.2 $\times$ 103 km
C
6.4 $\times$ 103 km
D
1.28 $\times$ 104 km

## Explanation

Energy required to move a satellite from earth surface to height h is,

E1 = Uh $-$ Usurface

= $-$ ${{GMm} \over {R + h}} - \left( { - {{GMm} \over R}} \right)$

= GMm $\left( {{1 \over R} - {1 \over {R + h}}} \right)$

= ${{GMm} \over {R(R + h)}} \times h$

We know, for sattelite at height h.

Centrifigual force = Gravitational force

$\Rightarrow$  ${{m{v^2}} \over {R + h}} = {{GMm} \over {{{\left( {R + h} \right)}^2}}}$

$\Rightarrow$  $mv$2 = ${{GMm} \over {R + h}}$

$\therefore$   ${1 \over 2}m{v^2}$ = ${{GMm} \over {2\left( {R + h} \right)}}$

$\therefore$  Kinetic energy (E2) = ${{GMm} \over {2(R + h)}}$

Given that,

E1 = E2

$\therefore$  ${{GMm} \over {R(R + h)}} \times h = {{GMm} \over {2\left( {R + h} \right)}}$

$\Rightarrow$  ${h \over R} = {1 \over 2}$

$\Rightarrow$  h = ${R \over 2}$

$\therefore$  h = ${{6.4 \times {{10}^3}} \over 2}$

= 3.2 $\times$ 103 km
2

### JEE Main 2019 (Online) 10th January Morning Slot

A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is -
A
mv2
B
${1 \over 2}$ mv2
C
${3 \over 2}$ mv2
D
2 mv2

## Explanation

At height r from center of earth. orbital velocity

= $\sqrt {{{GM} \over r}}$

$\therefore$  By energy conservation

KE of 'm' + $\left( { - {{GMm} \over r}} \right)$ = 0 + 0

(At infinity, PE = KE = 0)

$\Rightarrow$  KE of 'm' = ${{{GMm} \over r}}$ = ${\left( {\sqrt {{{GM} \over r}} } \right)^2}$ m = mv2
3

### JEE Main 2019 (Online) 10th January Evening Slot

Two stars of masses 3 $\times$ 1031 kg each, and at distance 2 $\times$ 1011 m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is - (Take Gravitational constant; G = 6.67 $\times$ 10–11 Nm2 kg–2)
A
2.4 $\times$ 104 m/s
B
1.4 $\times$ 105 m/s
C
3.8 $\times$ 104 m/s
D
2.8 $\times$ 105 m/s

## Explanation

By energy convervation between 0 & $\infty$.

$- {{GMm} \over r} + {{ - GMm} \over r} + {1 \over 2}m{V^2} = 0 + 0$

[M is mass of star m is mass of meteroite)

$\Rightarrow$ v $= \sqrt {{{4GM} \over r}} = 2.8 \times {10^5}$m/s
4

### JEE Main 2019 (Online) 11th January Morning Slot

A satellite is revolving in a circular orbit at a height h form the earth surface, such that h < < R where R is the earth. Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is :
A
$\sqrt {gR} \left( {\sqrt 2 - 1} \right)$
B
$\sqrt {2gR}$
C
$\sqrt {gR}$
D
${{\sqrt {gR} } \over 2}$

## Explanation

v0 = $\sqrt {g(R + h)} \approx \sqrt {gR}$

ve = $\sqrt {2g(R + h)} \approx \sqrt {2gR}$

$\Delta$v=ve $-$ v0 = $\left( {\sqrt 2 - 1} \right)\sqrt {gR}$