Question

Derive the expression for gravitational potential energy.

Answer 1

Consider two masses m₁ and m₂ are initially separated by a distance r’. m₁ is assumed to be fixed in its position.

Two distant masses changing the linear distance

To move the mass m₂ through an infinitesimal displacement `"d"vec"r"` from `vec"r"`  to `vec"r" + "d"vec"r"` (shown in the Figure above), work has to be done externally. This infinitesimal work is given by

dW = `vec"F"_"ext"."d"vec"r"` ......(1)

The work is done against the gravitational force, therefore,

`|vec"F"_"ext"| = |vec"F"_"G"| = ("Gm"_1"m"_2)/"r"^2` .........(2)

Substituting Equation (2) in (1) we get,

dW = `("Gm"_1"m"_2)/"r"^2 hat"r" . "d"vec"r"` ....(3)

And `"d"vec"r" = "dr"hat"r"` ....(4)

dW = `("Gm"_1"m"_2)/"r"^2hat"r".("dr"hat"r")` ......(5)

`hat"r".hat"r"` = 1 (since both are unit vectors)

∴ dW = `("Gm"_1"m"_2)/"r"^2 "dr"` ....(6)

Hence the total work done for displacing the particle from r’ to r is

W = `int_"r′"^"r""dW" = int_"r′"^"r"("Gm"_1"m"_2)/"r"^2 "dr"` ....(7)

W = `-(("Gm"_1"m"_2)/"r")_"r’"^"r"`

W = `-("Gm"_1"m"_2)/"r" + ("Gm"_1"m"_2)/"r’"` .....(8)

W = U(r) − U(r’)

where U(r) = `(-"Gm"_1"m"_2)/"r"`

This work done W gives the gravitational potential energy difference of the system of masses m₁ and m₂ when the separation between them is r and r’ respectively.

Cases for calculation of work done by gravity

Case 1:

If r < r’ - Since the gravitational force is attractive, m₂ is attracted by m₁. Then m₂ can move from r’ to r without any external Work. Here work is done by the system spending its internal energy and hence the work done is said to be negative.

Case 2:

If r > r’ - Work has to be done against gravity to move the object from r’ to r. Therefore work is done on the body by external force and hence work done is positive.