Question 199: Gravitation

Question

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 $$ 	imes $$ 10
3
km) is E
1
and kinetic energy required for the satellite to be in a circular orbit at this height is E
2
. The value of h for which E
1
and E
2
are equal, is

Accepted Answer

Energy required to move a satellite from earth surface to height h is,
E
1
= U
h
$$-$$ U
surface
= $$-$$ $${{GMm} \over {R + h}} - \left( { - {{GMm} \over R}} ight)$$
= GMm $$\left( {{1 \over R} - {1 \over {R + h}}} ight)$$
= $${{GMm} \over {R(R + h)}} 	imes h$$
We know, for sattelite at height h.
Centrifigual force = Gravitational force
$$ \Rightarrow $$  $${{m{v^2}} \over {R + h}} = {{GMm} \over {{{\left( {R + h} ight)}^2}}}$$
$$ \Rightarrow $$  $$mv$$
2
= $${{GMm} \over {R + h}}$$
$$ 	herefore $$   $${1 \over 2}m{v^2}$$ = $${{GMm} \over {2\left( {R + h} ight)}}$$
$$ 	herefore $$  Kinetic energy (E
2
) = $${{GMm} \over {2(R + h)}}$$
Given that,
E
1
= E
2
$$ 	herefore $$  $${{GMm} \over {R(R + h)}} 	imes h = {{GMm} \over {2\left( {R + h} ight)}}$$
$$ \Rightarrow $$  $${h \over R} = {1 \over 2}$$
$$ \Rightarrow $$  h = $${R \over 2}$$
$$ 	herefore $$  h = $${{6.4 	imes {{10}^3}} \over 2}$$
= 3.2 $$ 	imes $$ 10
3
km

Answer

1.6 $$ 	imes $$ 10
3
km

Answer

3.2 $$ 	imes $$ 10
3
km

Answer

6.4 $$ 	imes $$ 10
3
km

Answer

1.28 $$ 	imes $$ 10
4
km

Gravitation

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 $ \times $ 10 3 km) is E 1 and kinetic energy required for the satellite to be in a circular orbit at this height is E 2 . The value of h for which E 1 and E 2 are equal, is

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