Chapter 6: Problem 798
Kepler's second law regarding constancy of aerial velocity of a palnet is consequence of the law of conservation of (A) energy (B) angular Momentum (C) linear momentum (D) None of these
Chapter 6: Problem 798
Kepler's second law regarding constancy of aerial velocity of a palnet is consequence of the law of conservation of (A) energy (B) angular Momentum (C) linear momentum (D) None of these
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Get started for freeTwo satellites \(\mathrm{A}\) and \(\mathrm{B}\) go round a planet in circular orbits having radii \(4 \mathrm{R}\) and \(\mathrm{R}\) respectively If the speed of satellite \(\mathrm{A}\) is \(3 \mathrm{v}\), then speed of satellite \(\mathrm{B}\) is (A) \((3 \mathrm{v} / 2)\) (B) \((4 \mathrm{v} / 2)\) (C) \(6 \mathrm{v}\) (D) \(12 \mathrm{v}\)
There are two planets, the ratio of radius of two planets is \(\mathrm{k}\) but the acceleration due to gravity of both planets are \(\mathrm{g}\) what will be the ratio of their escape velocity. (A) \((\mathrm{kg})^{1 / 2}\) (B) \((\mathrm{kg})^{-1 / 2}\) (C) \((\mathrm{kg})^{2}\) (D) \((\mathrm{kg})^{-2}\)
Rockets are launched in eastward direction to take advantage of (A) the clear sky on eastern side (B) the thiner atmosphere on this side (C) earth's rotation (D) earth's tilt
Potential energy of a satellite having mass \(\mathrm{m}\) and rotating at a height of \(6.4 \times 10^{6} \mathrm{~m}\) from the surface of earth (A) \(-0.5 \mathrm{mg} \operatorname{Re}\) (B) \(-\mathrm{mg} \mathrm{Re}\) (C) \(-2 \mathrm{mgRe}\) (D) \(4 \mathrm{mgRe}\)
A body of mass \(\mathrm{m}\) is taken from earth surface to the height \(\mathrm{h}\) equal to radius of earth, the increase in potential energy will be (A) \(\operatorname{mg} R\) (B) \((1 / 2) \mathrm{mgR}\) (C) \(2 \mathrm{mg} \mathrm{R}\) (D) \((1 / 4) \mathrm{mgR}\)
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