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Chapter 6: Problem 719

In a gravitational field, at a point where the gravitational potential is zero (A) The gravitational field is necessarily zero (B) The gravitational field is not necessarily zero (C) Nothing can be said definitely, about the gravitational field (D) None of these

Short Answer

Expert verified
In a gravitational field, at a point where the gravitational potential is zero, the gravitational field is the negative derivative of the gravitational potential with respect to distance. As no information about the behavior of gravitational potential around that point is given, nothing can be said definitely about the gravitational field in this situation. Hence, the correct answer is (C) Nothing can be said definitely about the gravitational field.

Step by step solution

01

Gravitational Potential and Gravitational Field Formula

In a gravitational field, the gravitational potential (V) is the amount of work done per unit mass to bring an object from infinity to a certain point. The gravitational field (g) represents the force experienced by a unit mass placed at that point. Mathematically, these are represented as: \(V = -\dfrac{GM}{r}\) \(g = \dfrac{GM}{r^2}\) Where \(G\) is the gravitational constant, \(M\) is the mass of the object producing the gravitational field, and \(r\) is the distance from the center of the object.
02

Relationship Between Gravitational Potential and Gravitational Field

Now, we know that the gravitational potential is zero at the given point. We need to find out the behavior of the gravitational field at this point. To do so, let's differentiate the gravitational potential with respect to distance (\(r\)): \(\dfrac{dV}{dr} = \dfrac{d\left(-\dfrac{GM}{r}\right)}{dr} = \dfrac{GM}{r^2} \) \(g = -\dfrac{dV}{dr}\) This equation shows that the gravitational field is the negative derivative of the gravitational potential with respect to distance.
03

Evaluating the Gravitational Field in the Given Situation

According to the given exercise, the gravitational potential is zero at a certain point. It doesn't provide any information about the behavior of gravitational potential around that point. So, we cannot determine if the gravitational field is zero or nonzero at that point. To answer this question, we need more information about the behavior of gravitational potential around that point. Therefore, the correct answer is: (C) Nothing can be said definitely about the gravitational field.

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Most popular questions from this chapter

A planet is revolving round the sun in elliptical orbit. Velocity at perigee position (nearest) is \(\mathrm{v}_{1} \mid\) and at apogee position (farthest) is \(\mathrm{v}_{2}\) Both these velocities are perpendicular to the joining center of sun and planet \(r\) is the minimum distance and \(\mathrm{r}_{2}\) the maximum distance. (1) when the planet is at perigee position, it wants to revolve in a circular orbit by itself. For this value of \(\mathrm{G}\) (A) Should increase (B) Should decrease (C) data is insufficient (D) will not depend on the value of \(\mathrm{G}\) (2) At apogee position suppose speed of planer is slightly decreased from \(\mathrm{v}_{2}\), then what will happen to minimum distance \(r_{1}\) in the subsequent motion (A) \(r_{1}\) and \(r_{2}\) both will decreases (B) \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) both will increases (C) \(\mathrm{r}_{2}\) will remain as it is while \(\mathrm{r}_{1}\) will increase (D) \(\mathrm{r}_{2}\) will remain as it is while \(\mathrm{r}_{1}\) will decrease

If the height of a satellite from the earth is negligible in comparison of the radius of the earth \(\mathrm{R}\), the orbital velocity of the satellite is (A) \(\mathrm{gR}\) (B) \((\mathrm{gR} / 2)\) (C) \(\sqrt{(g} / \mathrm{R})\) (D) \(\sqrt{(g R)}\)

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

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 $\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$

A particle of mass \(10 \mathrm{~g}\) is kept on the surface of a uniform sphere of mass \(100 \mathrm{~kg}\) and radius \(10 \mathrm{~cm}\). Find the work to be done against the gravitational force between them to take the particle is away from the sphere \(\left(\mathrm{G}=6.67 \times 10^{-11} \mathrm{SI}\right.\) unit \()\) (A) \(6.67 \times 10^{-9} \mathrm{~J}\) (B) \(6.67 \times 10^{-10} \mathrm{~J}\) (C) \(13.34 \times 10^{-10} \mathrm{~J}\) (D) \(3.33 \times 10^{-10} \mathrm{~J}\)
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