JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 17)
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
1.6 $$ \times $$ 103 km
3.2 $$ \times $$ 103 km
6.4 $$ \times $$ 103 km
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
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
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