Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.

Question

Three masses $200\,\text{kg}$, $300\,\text{kg}$ and $400\,\text{kg}$ are placed at the vertices of an equilateral triangle of side $20\,\text{m}$. They are rearranged on the vertices of a bigger triangle of side $25\,\text{m}$ with the same centre. The work done in this process is ___ J.
(Gravitational constant $G = 6.7 \times 10^{-11}\,\text{N m}^2\text{kg}^{-2}$)

Answer 1

Statement	Correct Choice	Explanation
(a) Total energy is negative of kinetic/potential energy	kinetic	Total energy $ E = -\frac{GMm}{2r} = -KE $, since $ KE = \frac{GMm}{2r} $ and $ PE = -2KE $ for circular orbit.
(b) Energy to launch satellite vs stationary object	less	Orbiting satellite already has $ KE = \frac{GMm}{2r} $, needs only $ +\frac{GMm}{2r} $ more to reach $ E=0 $. Stationary object at same height needs full $ +\frac{GMm}{r} $.

Solution and Explanation

Top Questions on Gravitational Potential Energy

Questions Asked in CBSE Class XI exam

Statement	Correct Choice	Explanation
(a) Total energy is negative of kinetic/potential energy	kinetic	Total energy \( E = -\frac{GMm}{2r} = -KE \), since \( KE = \frac{GMm}{2r} \) and \( PE = -2KE \) for circular orbit.
(b) Energy to launch satellite vs stationary object	less	Orbiting satellite already has \( KE = \frac{GMm}{2r} \), needs only \( +\frac{GMm}{2r} \) more to reach \( E=0 \). Stationary object at same height needs full \( +\frac{GMm}{r} \).