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Chapter 6: Problem 687

A spherical planet far out in space has mass \(\mathrm{M}_{0}\) and diameter \(\mathrm{D}_{0}\). A particle of \(\mathrm{m}\) falling near the surface of this planet will experience an acceleration due to gravity which is equal to (A) $\left[\left(\mathrm{GM}_{0}\right) /\left(\mathrm{D}_{\circ}^{2}\right)\right]$ (B) $\left[\left(4 \mathrm{mGM}_{0}\right) /\left(\mathrm{D}_{0}^{2}\right)\right]$ (C) $\left[\left(4 \mathrm{GM}_{0}\right) /\left(\mathrm{D}_{0}^{2}\right)\right]$ (D) $\left[\left(\mathrm{GmM}_{0}\right) /\left(\mathrm{D}_{\circ}^{2}\right)\right]$

Short Answer

Expert verified
The acceleration due to gravity acting on a particle of mass m near the surface of a spherical planet with mass M0 and diameter D0 can be found using the gravitational force formula and simplification. The acceleration due to gravity (g) is given by: \[\boxed{g = \frac{4GM_0}{D_0^2}}\] This corresponds to option (C).

Step by step solution

01

Understand the gravitational force formula

The gravitational force between two objects is given by Newton's Law of Universal Gravitation: \[F = G \frac{m_1 m_2}{r^2}\] Where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between the centers of mass of the two objects.
02

Rewrite the formula to find the acceleration due to gravity

We can rewrite the gravitational force formula by solving for the acceleration due to gravity acting on the particle (g) instead of the gravitational force. Since F = m * g, we have: \[g = \frac{F}{m} = G \frac{m_1 m_2}{m r^2}\] We can now substitute the planet's mass (M0) and particle's mass (m) and the distance between their centers of mass (D0): \[g = G \frac{M_0 m}{r^2}\]
03

Calculate the distance between the centers of masses

Since the particle is near the surface of the planet, the distance between their centers of mass (r) is approximately equal to the radius of the planet. Since diameter D0 = 2 * radius, we can find the radius: \[r = \frac{D_0}{2}\]
04

Substitute the distance into the equation for acceleration due to gravity

Now, let's substitute the found distance (r) into the equation for acceleration due to gravity (g): \[g = G \frac{M_0 m}{\left(\frac{D_0}{2}\right)^2} = G \frac{M_0 m}{\frac{D_0^2}{4}}\]
05

Simplify the equation

By simplifying the equation, we can cancel out the mass m of the particle and get the expression for the acceleration due to gravity only in terms of the gravitational constant (G), mass of the planet (M0), and diameter of the planet (D0): \[g = \frac{4GM_0}{D_0^2}\]
06

Compare with the given options

Comparing our found expression for the acceleration due to gravity (g) with the given options, we see that it corresponds to option (C): \[\boxed{g = \left[\left(4GM_0\right)/\left(D_0^2\right)\right]}\]

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Most popular questions from this chapter

A satellite revolves around the earth in an elliptical orbit. Its speed (A) is the same at all points in the orbit (B) is greatest when it is closest to the earth (C) is greatest when it is farthest to the earth (D) goes on increasing or decreasing continuously depending upon the mass of the satellite

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) \((1 / 4) \mathrm{mgR}\)

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) \(4 \mathrm{mgRe}\)

When a particle is projected from the surface of earth, it mechanical energy and angular momentum about center of earth at all time is constant (i) A particle of mass \(\mathrm{m}\) is projected from the surface of earth with velocity \(\mathrm{V}_{0}\) at angle \(\theta\) with horizontal suppose \(\mathrm{h}\) be the maximum height of particle from surface of earth and \(\mathrm{v}\) its speed at that point them \(\mathrm{V}\) is (A) \(\mathrm{V}_{0} \cos \theta\) \((\mathrm{B})>\mathrm{V}_{0} \cos \theta\) (C) \(<\mathrm{V}_{0} \cos \theta\) (D) zero (ii) Maximum height h of the particle is $(\mathrm{A})=\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$ (B) $>\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$ $(\mathrm{C})<\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$ (D) can be greater than or less than $\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$

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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