For a body in a circular orbit around Earth of radius r, the ratio of kinetic energy to total mechanical energy is
A{'text': '1 : 1', 'label': 'A'}
B{'text': 'Undefined', 'label': 'B'}
C{'text': '2 : (−1)', 'label': 'C'}
D{'text': '1 : (−1)', 'label': 'D'}
Answer & Solution
Correct answer: D. {'text': '1 : (−1)', 'label': 'D'}
1. For a circular orbit: KE = (1/2) m v² = G M m / (2 r).
2. PE = −G M m / r; total energy E = KE + PE = −G M m / (2 r).
3. Compare: KE = −E, so KE / E = −1, or equivalently 1 : (−1).
4. This is the standard virial-theorem relation for gravitational orbits.
_Source: NCERT Class 11 Physics, Ch 7 "Gravitation", §7.7_
Related questions
Cavendish's torsion balance experiment (1798) measuredThe gravitational field near Earth's surface is approximately uniform becauseNewton showed that the same law that makes an apple fall to the ground alsoThe weight of an object at a certain height above Earth's surface equals its weight at depThe variation of g with depth d below Earth's surface (assuming uniform density) isA geostationary satellite orbits Earth at a height where its period equalsThe gravitational potential energy of two point masses m₁ and m₂ separated by distance r iAn orbiting satellite has an orbital speed v = √(G M / r). If the orbital radius r is quad