Question

Two lead spheres of same radii are in contact with each other. The gravitational force of attraction between them is F. If two lead spheres of double the previous radii are in contact with each other, the gravitational force of attraction between them now will be

Answer 1

Answer:

The force will be 16F

Explanation:

If the radius of the lead sphere is R and mass M then the gravitational force when they are in contact is given by

$F=\frac{GM^2}{(2R)^2}$

$\implies F=\frac{GM^2}{4R^2}$

When the radii of the lead spheres are doubled, since the density is same for lead, the mass will be changed

If the density is assumed to be d then

Previously,

$M=d\times V$

$\implies M=d\times \frac{4}{3}\pi R^3$

Now the mass will be

$M'=d\times\frac{4}{3}\pi\times (2R)^3$

$\implies M'=d\times\frac{4}{3}\times 8R^3$

$\implies M'=8M$

Therefore, the new gravitational force between them will be

$F'=\frac{GM'^2}{(4R)^2}$

$\implies F'=\frac{G\times (8M)^2}{16R^2}$

$\implies F'=\frac{GM^2\times 64}{16R^2}$

$\implies F'=16\times\frac{GM^2}{4R^2}$

$\implies F'=16F$

Therefore, the force between the lead spheres will now be 16F

Hope this helps.