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Relation Between Gravitational Field And Potential - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • 3 Questions around this concept.

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In a certain region of space gravitational field is given by

\mathrm{ I=-\left(\frac{k}{r}\right)}. Taking the reference point to be at \mathrm{r=r_0 \: with \: v=v_0,} the potential \mathrm{ (V)} is given by
 

\mathrm{P} is a point at a distance \mathrm{r} from the centre of a solid sphere of radius \mathrm{R_0}. The gravitational potential at \mathrm{P \: is \: V}. If  \mathrm{ V} is plotted as a function of \mathrm{r}, then the curve representing the plot correctly is

The gravitational field due to a mass distribution is given by \mathrm{I=\left(A / x^3\right)}  in X−direction. The gravitational potential at a distance x is equal to

Concepts Covered - 1

Relation between gravitational field and potential

 

Gravitational field and potential are related as

\vec{E}=-\frac{dV}{dr}

Where E is Gravitational field

And V is Gravitational potential

And r is the position vector 

And Negative sign indicates that in the direction of intensity the potential decreases.

 If \ \ \vec{r}=x \ \vec{i}+y \ \vec{j}+z \ \vec{k}

 

Then E_x=\frac{\delta V}{dx},E_y=\frac{\delta V}{dy},E_z=\frac{\delta V}{dz}

Proof-

Let gravitational field at a point r due to a given mass distribution is E.

If a test mass m is placed inside a uniform gravitational field E.

Then force on a particle m when it is at r is  \vec{F}=m\vec{E} as shown in figure

      

 

As the particle is displaced from r to r + dr the

work done by the gravitational force on it is

 

dW=\vec{F}.\vec{r}=m\vec{E}.d\vec{r}

The change in potential energy during this

displacement is

dU=-dW=-\vec{F}.\vec{r}=-m\vec{E}.d\vec{r}

And we know that Relation between Potential and Potential energy

As U=mV

So dV=\frac{dU}{m}= -\vec{E}.d\vec{r}

Integrating between r1, and r2

We get V(\vec{r_2})-V(\vec{r_1})= \int_{r_1}^{r_2} -\vec{E}.d\vec{r}

If r1=r0, is taken at the reference point, V(r0) = 0. 

Then the potential V(r2=r) at any point r is 

V(\vec{r}) = \int_{r_0}^{r} -\vec{E}.d\vec{r}

in Cartesian coordinates, we can write

\vec{E}=E_x \ \vec{i}+E_y \ \vec{j}+E_z \ \vec{k}

If \ \ \vec{r}=x \ \vec{i}+y \ \vec{j}+z \ \vec{k}

Then d\vec{r}=dx \ \vec{i}+dy \ \vec{j}+dz \ \vec{k}

So  

\vec{E}.d\vec{r}=-dV=E_xdx +E_ydy +E_zdz\\ dV=-E_xdx -E_ydy -E_zdz

If y and z remain constant, dy = dz = 0

Thus E_x=\frac{dV}{dx}

Similarly E_y=\frac{dV}{dy}, E_z=\frac{dV}{dz}

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Relation between gravitational field and potential

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