Chapter 6: Problem 703
Weight of a body is maximum at (A) moon (B) poles of earth (C) Equator of earth (D) Center of earth
Chapter 6: Problem 703
Weight of a body is maximum at (A) moon (B) poles of earth (C) Equator of earth (D) Center of earth
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Get started for freeThe Gravitational P.E. of a body of mass \(\mathrm{m}\) at the earth's surface is \(-\mathrm{mgRe}\). Its gravitational potential energy at a height \(\operatorname{Re}\) from the earth's surface will be \(=\ldots \ldots \ldots\) here (Re is the radius of the earth) (A) \(-2 \mathrm{mgRe}\) (B) \(2 \mathrm{mgRe}\) (C) \((1 / 2) \mathrm{mg} \mathrm{Re}\) (D) \(-(1 / 2) \mathrm{mg} \operatorname{Re}\)
Direction (Read the following questions and choose) (A) If both Assertion and Reason are true and the Reason is correct explanation of assertion (B) If both Assertion and Reason are true, but reason is not correct explanation of the Assertion (C) If Assertion is true, but the Reason is false (D) If Assertion is false, but the Reason is true Assertion: The square of the period of revolution of a planet is proportional to the cube of its distance from the sun. Reason: Sun's gravitational field is inversely proportional to the square of its distance from the planet (a) \(\mathrm{A}\) (b) \(\mathrm{B}\) (c) \(\mathrm{C}\) (d) D
Given mass of the moon is \((1718)\) of the mass of the earth and corresponding radius is \((1 / 4)\) of the earth, If escape velocity on the earth surface is \(11.2 \mathrm{kms}^{-1}\) the value of same on the surface of moon is $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$. (A) \(0.14\) (B) \(0.5\) (C) \(2.5\) (D) 5
The additional K.E. to be provided to a satellite of mass \(\mathrm{m}\) revolving around a planet of mass \(\mathrm{M}\), to transfer it from a circular orbit of radius \(\mathrm{R}_{1}\) to another radius \(\mathrm{R}_{2}\left(\mathrm{R}_{2}>\mathrm{R}_{1}\right)\) is (A) $\operatorname{GMm}\left[\left(1 / R_{1}^{2}\right)-\left(1 / R_{2}^{2}\right)\right]$ \(\operatorname{GMm}\left[\left(1 / R_{1}\right)-\left(1 / R_{2}\right)\right]\) (C) $2 \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$ (D) $(1 / 2) \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$
The escape velocity of an object from the earth depends upon the mass of earth (M), its mean density ( \(p\) ), its radius (R) and gravitational constant (G), thus the formula for escape velocity is (A) \(U=\mathrm{R} \sqrt{[}(8 \pi / 3) \mathrm{Gp}]\) (C) \(\mathrm{U}=\sqrt{(2 \mathrm{GMR})}\) (D) \(U=\sqrt{\left[(2 \mathrm{GMR}) / \mathrm{R}^{2}\right]}\)
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