Name: Understanding Gravitational Field
Uploaded: 2019-07-19
Duration: 5 min
Description: Each mass exerts a force on the other mass. This phenomenon is explained using the gravitational field. Let's watch this video to understand the gravitationa...

Question

Find the gravitational potential

(V)

and intensity

(E)

at a point at a distance

r

from the centre of a hollow sphere of radius

R

, where

r

varies from

0

to

\infty

.

Accepted Answer

Consider a thin uniform hollow sphere of radius $R$ as in the figure. We have to find the gravitational potential and the intensity of the hollow sphere.

The mass of the hollow sphere be $r$ . Consider a point $R$ at $r$ distance away from the centre of the sphere. Let $0$ . We need to find the gravitational potential of the hollow sphere at that point.

To make it easier, let the sphere be a football or an onion (but hollow). Consider cutting an onion into many slices, with each slice having very small width. So the full onion can be treated as those infinite slices of onion put together. But, remember the onion was a hollow one. So the slices will be a ring of infinitesimal width.
From the figure, consider only a slice of the sphere (or the onion) $A B C D$ . It is evident that the radius and width of the slice (ring) is $R \sin (θ)$ and $R d θ$ respectively. The point $P$ will be $x$ distance away from the slice.
Therefore, the area of the slice will be $2 π (radius) (width of the slice)$

Area of each slice will be, $d A = 2 π (R s i n θ) (R d θ) = 2 π (R^{2} s i n θ d θ)$
Since it is a uniform sphere, the mass of the slice dm will be the product of area and density. $d m = 2 π σ (R^{2} s i n θ d θ)$
Thus, the gravitational potential at $P$ will be,

$d V = - \frac{G d m}{x} d V = - \frac{G 2 π σ (R^{2} s i n θ d θ)}{x}$

From the figure, using some vector identities, it could be possible to eliminate the angular dependence $θ$

From $∆ O A P,$ $\vec{R} + \vec{x} = \vec{r} \Rightarrow x^{2} = r^{2} + R^{2} - 2 R r \cos θ$

$x d x = R r \sin θ d θ \Rightarrow \frac{x d x}{R r} = \sin θ d θ$

Therefore, $d V = - \frac{G 2 π σ (R^{2} s i n θ d θ)}{x} \Rightarrow d V = - \frac{G 2 π σ R d x}{r}$

Now, for the full contribution, the slices will be from $F$ to $G$ . So looking from $P$ the distance $x$ varies from $r - R$ to $r + R$ ,

Integrating,

$V = - \int_{r - R}^{r + R} \frac{G 2 π σ R d x}{r} V = - \frac{G 2 π σ R (r + R - (r - R))}{r} V = - \frac{G 4 π σ R^{2}}{r} = - \frac{G m}{r}$

The gravitational intensity at $P$ can be expressed immediately from the above expression, $E = - \frac{d V}{d r} \overset{⏜}{r}$
$E = - \frac{G m}{r^{2}} \overset{⏜}{r}$

For the second case, $r < R$ Inside the sphere, the integration limit will change as the point $P$ is inside the sphere. Starting from the above equation, $d V = - \frac{G 2 π σ R d x}{r}$ integrate from $R - r$ to $R + r$ in the same logic as above,

$V = - \int_{R - r}^{R + r} \frac{G 2 π σ R d x}{r} V = - \frac{G 2 π σ R (R + r - (R - r))}{r} V = - G 4 π σ R V = - G \frac{4 π σ R^{2}}{R} = - \frac{G m}{R}$

Now the intensity, as we know is the derivative of the potential with respect to the distance, called the gradient, the intensity will be,

$E = - \frac{d V}{d r} \overset{⏜}{r} \Rightarrow E = 0$

Find the gravitational potential $(V)$ and intensity $(E)$ at a point at a distance $r$ from the centre of a hollow sphere of radius $R$ , where $r$ varies from $0$ to $\infty$ .

Important Questions on Gravitation

Learn and Prepare for any exam you want !

Find the gravitational potential V and intensity E at a point at a distance r from the centre of a hollow sphere of radius R, where r varies from 0 to ∞.

Important Questions on Gravitation

Learn and Prepare for any exam you want !

Find the gravitational potential $(V)$ and intensity $(E)$ at a point at a distance $r$ from the centre of a hollow sphere of radius $R$ , where $r$ varies from $0$ to $\infty$ .