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Chapter 6: Problem 742

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) \((\mathrm{kg})^{-2}\)

Short Answer

Expert verified
The short answer is: (A) \((kg)^{1/2}\).

Step by step solution

01

Write down the escape velocity formula

The escape velocity formula is given by: \[v_e = \sqrt{\frac{2GM}{R}}\] where \(v_e\) is the escape velocity, \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(R\) is the radius of the planet.
02

Find the ratio of masses of the planets

We are given that the ratio of the radii of the planets is \(k\). We know that the acceleration due to gravity on both planets is the same, so we can write: \[g = \frac{GM_1}{R_1^2} = \frac{GM_2}{R_2^2}\] Dividing the equations and using the given ratio of radii, we get: \[\frac{M_1}{M_2} = \frac{R_1^2}{R_2^2} = k^2\]
03

Write down the escape velocities for both planets

Using the escape velocity formula and the ratio of masses found in the previous step, we can write the escape velocities for both planets as: \[v_{e1} = \sqrt{\frac{2GM_1}{R_1}}\] \[v_{e2} = \sqrt{\frac{2GM_2}{R_2}}\]
04

Find the ratio of escape velocities

Now, we will find the ratio of escape velocities by dividing the equations: \[\frac{v_{e1}}{v_{e2}} = \frac{\sqrt{\frac{2GM_1}{R_1}}}{\sqrt{\frac{2GM_2}{R_2}}}\] Using the ratios found in Step 2, we can simplify this expression: \[\frac{v_{e1}}{v_{e2}} = \sqrt{\frac{M_1}{M_2} \cdot \frac{R_2}{R_1}} = \sqrt{\frac{k^2}{k}} = \sqrt{k}\]
05

Identify the correct answer

Comparing our final answer to the provided options, we find that the correct answer is: (A) \((kg)^{1/2}\).

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Most popular questions from this chapter

The acceleration of a body due to the attraction of the earth (radius R) at a distance \(2 \mathrm{R}\) from the surface of the earth is \(=\) (g \(=\overline{\text { acceleration due to gravity at the surface of earth })}\) (A) \(\mathrm{g} / 9\) (B) \(\mathrm{g} / 3\) (C) \(\mathrm{g} / 4\) (D) 9

The largest and shortest distance of the earth from the sun are \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) its distance from the sun when it is at the perpendicular to the major axis of the orbit drawn from the sun (A) \(\left[\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right) / 4\right]\) (B) $\left[\left(\mathrm{r}_{1} \mathrm{r}_{2}\right) /\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right)\right]$ (C) \(\left[\left(2 r_{1} r_{2}\right) /\left(r_{1}+r_{2}\right)\right]\) (D) \(\left[\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right) / 3\right]\)

A satellite of mass \(\mathrm{m}\) is orbiting close to the surface of the earth (Radius \(\mathrm{R}=6400 \mathrm{~km}\) ) has a K.E. \(\mathrm{K}\). The corresponding \(\mathrm{K} . \mathrm{E}\). of satellite to escape from the earth's gravitational field is (A) \(\mathrm{K}\) (B) \(2 \mathrm{~K}\) (C) \(\mathrm{mg} \mathrm{R}\) (D) \(\mathrm{m} \mathrm{K}\)

An artificial satellite moving in a circular orbit around earth has a total (kinetic + potential energy) \(E_{0}\), its potential energy is (A) \(-\mathrm{E}_{0}\) (B) \(1.5 \mathrm{E}_{0}\) (C) \(2 \mathrm{E}_{0}\) (D) \(\mathrm{E}_{0}\)

Gravitational potential at any point inside a spherical shall is uniform and is given by \(-(\mathrm{GM} / \mathrm{R})\) where \(\mathrm{M}\) is the mass of shell and \(\mathrm{R}\) its radius. At the center solid sphere, potential is \([-\\{(3 \mathrm{GM}) /(2 \mathrm{R})\\}]\)(1) There is a concentric hole of radius \(\mathrm{R}\) in a solid sphere of radius \(2 \mathrm{R}\) mass of the remaining portion is \(\mathrm{M}\) what is the gravitational at center? (A) \(-[(3 \mathrm{GM}) / 7 \mathrm{R}]\) (B) \(-[(5 \mathrm{GM}) / 7 \mathrm{R}]\) (C) \(-[(7 \mathrm{GM}) / 14]\) (D) \(-[(9 \mathrm{GM}) /(14 \mathrm{R})]\)
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