Chapter 6: Problem 789
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
Chapter 6: Problem 789
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
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Get started for freeSuppose the gravitational force varies inversely as the nth power of distance the time period of planet in circular orbit of radius \(\mathrm{R}\) around the sun will be proportional to (A) \(\mathrm{R}^{[(\mathrm{n}+1) / 2]}\) (B) \(\mathrm{R}^{[(\mathrm{n}-1) / 2]}\) (C) \(\mathrm{R}^{\mathrm{n}}\) (D) \(\mathrm{R}^{[(\mathrm{n}-1) / 2]}\)
The escape velocity for a body projected vertically upwards from the surface of earth is \(11 \mathrm{kms}^{-1}\). If the body is projected at an angle of \(45^{\circ}\) with the vertical, the escape velocity will be $\ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \((11 / \sqrt{2})\) (B) \(11 \sqrt{2}\) (C) 22 (D) 11
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 moon's radius is \(1 / 4\) that of earth and its mass is \(1 / 80\) times that of the earth. If g represents the acceleration due to gravity on the surface of earth, that on the surface of the moon is (A) \(g / 4\) (B) \(\mathrm{g} / 5\) (c) \(\mathrm{g} / 6\) (D) \(\mathrm{g} / 8\)
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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