Question: 7 Two bodies of mass m and 9 m are placed at a distance R. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ( G= gravitational constant) :
(1) -\frac{16 G m}{R}
(2) -\frac{20 G m}{R}
(3) -\frac{8 G m}{R}
(4) -\frac{12 G m}{R}
Answer: Option (1)
Explanation:
Let the point where gravitational field is zero be at a distance x from the mass
m and hence at a distance R-x from the mass 9m.
At this point, the magnitudes of gravitational fields due to both masses are equal.
So, \frac{Gm}{x^{2}}=\frac{G(9m)}{(R-x)^{2}}.
Canceling Gm from both sides,
we get \frac{1}{x^{2}}=\frac{9}{(R-x)^{2}}.
Taking square root, \frac{1}{x}=\frac{3}{R-x}.
Solving, R-x=3x, hence x=\frac{R}{4}.
The gravitational potential at this point is the sum of potentials due to both masses.
V=-\frac{Gm}{x}-\frac{G(9m)}{R-x}.
Substituting x=\frac{R}{4} and R-x=\frac{3R}{4}, we get
V=-\frac{Gm}{R/4}-\frac{9Gm}{3R/4}.
Simplifying, V=-\frac{4Gm}{R}-\frac{12Gm}{R}.
Thus, V=-\frac{16Gm}{R}.
Therefore, the gravitational potential at the point
where the field is zero is -\frac{16 G m}{R}.