Question

Suppose the gravitational potential due to a small system is k/r2 at a distance r from it. What will be the gravitational field? Can you think of any such system? What happens if there were negative masses?

Answer 1

Explanation:

• It is given, the gravitational potential as $V=\frac{k}{r^{2}}$  where r is the distance from the system.  
• (a) The gravitational field due to the system can be defined as $E=-\frac{d v}{d r}$.
• (b) This kind of system is not possible.  On substituting the ‘V’ value in the above equation, we get  
• $E=\frac{2 k}{r^{8}}$.  Here, E is directly proportional to $\left(\frac{1}{r^{3}}\right)$, i.e. the gravitational field is proportional to inverse of cube of the distance.  But this is not experimentally true as $F_g$ is always proportional to inverse of square of the distance from the system.  
• (c) The system will be a dipole of two masses, i.e. a small distance separates the positive and negative mass.  
• At this scenario, the E can be directly proportional to $\left(\frac{1}{r^{3}}\right)$.  Still, this kind of system is possible only if there were negative masses to exist.

Answer 2

Gravitational field

Explanation:

The gravitational potential due to the system V = k/r²

Gravitational field due to the system E = -dV/dr

Substituting, we get E = -d(k/r²) / dr = 2 (-2k/r³)

So we find that for this system E ∝ 1/r³. E is inversely proportional to r.

This not possible because Fg is always proportional to inverse of square of distance from experiments.

This type of system would be possible only if there is negative mass.

This system is a dipole of two masses, where one mass is positive and other is negative, separated by a distance.

If there negative mass, the gravitational field due to the dipole is proportional to 1/r³.