Discussion
If the magnitude of gravitational potential energy of an object of mass 200 kg at a height of \(3.6 \times 10^{6}m\) from the earth surface is \(6 \times 10^{6}J\) then its value at a height of \(5.6 \times 10^{6}m\) is (Radius of earth is \(6.4 \times 10^{6}m\))
Answer: A
The potential energy at the new height has magnitude \(5 \times 10^{6}\) J.
The magnitude of gravitational potential energy is \(\frac{GMm}{r}\), with \(r\) measured from the centre of the earth, so it is inversely proportional to \(r\).
\[ \begin{aligned} r_{1} &= 6.4 + 3.6 = 10 \times 10^{6} \text{ m} \\ r_{2} &= 6.4 + 5.6 = 12 \times 10^{6} \text{ m} \\ U_{2} &= 6 \times 10^{6} \times \frac{10}{12} = 5 \times 10^{6} \text{ J} \end{aligned} \]
Using the heights above the surface instead of the distances from the centre gives \(6 \times 10^{6} \times \frac{3.6}{5.6}\), about \(3.9 \times 10^{6}\) J, which matches none of the options.
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