Question

A satellite of mass $1000 kg$ is supposed to orbit the earth at a height of $2000 km$ above the Earth's surface. Find,

$(a)$ Its speed in the orbit,

$(b)$ Its kinetic energy,

$(c)$ The potential energy of the earth-satellite system and

$(d)$ Its time period. Mass of the Earth $= 6 \times 10^{24} kg$ .

Accepted Answer

$(a)$ The orbital velocity of the satellite is $V_{o} = \sqrt{\frac{G M_{e}}{r}}$

where,

$M =$ mass of the Earth,

$r =$ distance of the satellite from the centre of Earth, which in this case is the sum of the radius of the Earth and the height above the surface,

$= 6400 + 2000 = 8400 km$

$V_{o} = \sqrt{\frac{(6.67 \times 10^{- 11}) (6 \times 10^{24})}{(8400000)}} m s^{- 1}$

$= 6.9 km s^{- 1}$

The orbital velocity of the satellite is $6.9 km s^{- 1}$ .

$(b)$ The kinetic energy of the satellite is $K E = \frac{1}{2} m v^{2}$

The velocity is equal to the orbital velocity and the mass is equal to $1000 kg$ .

$= \frac{1}{2} (1000) {(6.9 \times 10^{3})}^{2}$

$= 2.38 \times 10^{10} J$

$(c)$ The potential energy of the earth-satellite system is,

$U = - \frac{G M_{e} m}{r}$

( $M_{e} =$ mass of Earth which is $6 \times 10^{24} kg$ , $m =$ mass of satellite is $1000 kg$ )

$= - \frac{(6.67 \times 10^{- 11}) (6 \times 10^{24}) (1000)}{(8400) \times 10^{3}}$

$= - 4.76 \times 10^{10} J$

$(d)$ The time period can be calculated from the orbital velocity as,

$T = \frac{2 π r}{V_{o}} = \frac{2 π (8400) \times 10^{3}}{6.9 \times 10^{3}} = 7645.22 s$ $= 2.1 hours$ .

H C Verma Solutions for Chapter: Gravitation, Exercise 4: EXERCISES

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