Question

The gravitational P.E. of a rocket of mass 100 kg at a distance of ${10}^7$ m from the earths centre is $-4\times {10}^9$ J. The weight of the rocket at a distance of ${10}^9$ m from the centre of the earth is :

• A. $4\times {10}^{-2}$ N
• B. $4\times {10}^{-9}$ N
• C. $4\times {10}^{-6}$ N
• D. $4\times {10}^{-3}$ N

Answer 1

Answer: (A)

$4\times {10}^{-2}$ N

Gravitational potential energy is given by
$PE=-\frac{GMm}{R}=-4\times {10}^9$
But $R={10}^7$ and $PE=-4\times {10}^9$ J
$\Rightarrow GMm=4\times {10}^9\times {10}^7=4\times {10}^{16}$
Now the weight of the satellite at a radius of ${10}^{9}m$ from earth's centre is given by,
$mg=\frac{GMm}{R^2}=\frac{4\times {10}^{16}}{{10}^{18}}=4\times {10}^{-2}$ N