Question

Gravtational potential at a distance 'r' from a point mass 'm' is V = -(GM)/(r) Find gravitational field strength at that point.

Answer 1

Explanation:

Gravitational potential at any inside point is given as

V=−

2R

3

GM

(3R

2

−r

2

)....(i)

for r=

2

R

V=−

8R

11GM

Subtracting potential due to cavity of mass M

c

=

8

M

and R

c

=

2

R

Gravitational potential at center is obtained by substituting r=0 in equation (i) =−

2R

c

3GM

c

V=−

8R

11GM

−(−

2R

c

3GM

c

)=−

8R

11GM

+

2

2

R

3G

8

M

⇒V=−

R

GM

Answer 2

Given :

Gravitational potential at a distance 'r' = V

To find :

Gravitational field strength

Solution :

The gravitational field strength at any point r is the gravitational force exerted per unit mass placed on that point.

So,

$</strong></p><p><strong>[tex]\textbf{\large Gravitational field strength E =}-\frac{\textbf{\large dV}}{\textbf{\large dr}} \:$

(equation 1)

Here

V = Gravitational potential.

It is given here that

Gravitational potential V :

$V = \frac {-GM}{r} \:$

(equation 2)

So according to equation 1 when we differentiate Gravitational potential with respect to distance, we can get Gravitational field strength.

So

Putting value of equation 2 in equation 1 we get :

Gravitational field strength E :

$E = -\frac{d}{dr}(\frac{-GM}{r}) \:$

$E =GM\frac{d}{dr}(\frac{1}{r})$

So,

Gravitational field strength E :

$\textbf{ \Large E } = - \frac{ \textbf{ \Large GM}}{ \textbf{ \Large {r}}^{ \bf \: 2} }$