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Chapter 6: Problem 832

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) \(-[(9 \mathrm{GM}) /(14 \mathrm{R})]\)

Short Answer

Expert verified
The gravitational potential at the center of the sphere with a concentric hole is \(-\frac{5GM}{14R}\). The correct answer is (B) \(-[\frac{5GM}{7R}]\).

Step by step solution

01

Understand the Sphere Structure and Given Values

We have a solid sphere of radius 2R with a concentric hole of radius R. The mass of the remaining portion is M. We need to find the gravitational potential at the center.
02

Calculate the Mass of the Removed Portion of Sphere

To find the mass of the removed portion, we can equate the densities of the removed portion and the remaining portion since they are concentric: \(\rho_1 = \rho_2\) \(\frac{M_1}{\frac{4}{3}\pi R^3} = \frac{M-M_1}{\frac{4}{3}\pi (2R)^3 - \frac{4}{3}\pi R^3}\) Solving for \(M_1\), we get: \(M_1 = \frac{M}{7}\)
03

Calculate the Gravitational Potential of the Removed Portion

Using the given formula for the gravitational potential at the center of a solid sphere, we can find the potential due to the removed portion: \(V_1 = -\frac{3GM_1}{2R}\) Replacing \(M_1\) with the value obtained in Step 2, we get: \(V_1 = -\frac{3G(\frac{M}{7})}{2R}\)
04

Calculate the Gravitational Potential of the Whole Sphere

Now, let's find the gravitational potential at the center of the whole sphere of radius 2R and mass M: \(V_2 = -\frac{3GM}{2(2R)}\)
05

Find the Gravitational Potential at the Center of the Sphere with the Hole

We can find the gravitational potential at the center by subtracting the potential due to the removed portion from the potential of the whole sphere: \(V = V_2 - V_1\) Substituting the values of \(V_1\) and \(V_2\) from Steps 3 and 4, we get: \(V = -\frac{3GM}{4R} + \frac{3GM}{14R}\)
06

Simplify the Expression for Gravitational Potential

Combining the terms, we get the gravitational potential at the center: \(V = -\frac{5GM}{14R}\) Now, comparing with the given options, we see that the correct answer is (B) \(-[(5 \mathrm{GM}) / 7 \mathrm{R}]\).

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Most popular questions from this chapter

A body weights \(700 \mathrm{~g} \mathrm{wt}\) on the surface of earth. How much it weight on the surface of planet whose mass is \(1 / 7\) and radius is half that of the earth (A) \(200 \mathrm{~g} \mathrm{wt}\) (B) \(400 \mathrm{~g} \mathrm{wt}\) (C) \(50 \mathrm{~g} \mathrm{wt}\) (D) \(300 \mathrm{~g}\) wt.

According to keplar, the period of revolution of a planet ( \(\mathrm{T}\) ) and its mean distance from the sun (r) are related by the equation (A) \(\mathrm{T}^{3} \mathrm{r}^{3}=\) constant (B) \(\mathrm{T}^{2} \mathrm{r}^{-3}=\) constant (C) \(\mathrm{Tr}^{3}=\) constant (D) \(\mathrm{T}^{2} \mathrm{r}=\) constant

The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is (A) \(\sqrt{3} \mathrm{~V}_{\mathrm{e}}\) (B) \(3 \mathrm{~V}_{\mathrm{e}}\) (C) \(\sqrt{2} \mathrm{~V}_{\mathrm{e}}\) (D) \(2 \mathrm{~V}_{\mathrm{e}}\)

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) \(\mathrm{E}_{0}\)

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