Problem 2 : The gravitational field due to a mass distribution is given by the relation,
E
=
A
x
2
E
=
A
x
2
Find gravitational potential at “x”.
Solution : Gravitational field is equal to negative of first differential with respect to displacement in a given direction.
E
=

ⅆ
V
ⅆ
x
E
=

ⅆ
V
ⅆ
x
Substituting the given expression for “E”, we have :
⇒
A
x
2
=

ⅆ
V
ⅆ
x
⇒
A
x
2
=

ⅆ
V
ⅆ
x
⇒
ⅆ
V
=
−
A
ⅆ
x
x
2
⇒
ⅆ
V
=
−
A
ⅆ
x
x
2
Integrating between initial and final values of infinity and “x”,
⇒
Δ
V
=
V
f
−
V
i
=

A
∫
ⅆ
x
x
2
⇒
Δ
V
=
V
f
−
V
i
=

A
∫
ⅆ
x
x
2
We know that potential at infinity is zero gravitational potential reference. Hence,
V
i
V
i
= 0. Let
V
f
V
f
= V, then:
⇒
V
=

A
[

1
/
x
]
∞
x
=

A
[

1
x
+
0
]
=
A
x
⇒
V
=

A
[

1
/
x
]
∞
x
=

A
[

1
x
+
0
]
=
A
x