Question

A solid lead sphere of radius 10 m (about 66 feet across!) has a mass of about 57 million kg. If two of these spheres are floating right next to each other (centers 20 m apart) in deep space, the gravitational attraction between the spheres is only 540 N (about 100 pounds). How large would this gravitational force be if the distance between the centers of the two spheres were doubled????

please and thank you!!

Answer 1

Gravitational force = G * M₁ * M₂ / R² = 6.67 * 10^-11 * 57 * 10⁶ * 57 * 10⁶ / 40² = 13.54 *10 = 135.4 N

Answer 2

Force is inversely proportional to square of distance.

If distance is doubled, force will become 1/4 th.

Force = 540 /4 = 135 N

Answer 3

To do this use this equation

F = Gm1m2/ r^2
G = 6.6726 x 10^-11N-m2/kg2

F = [(6.6726 x 10^-11)(57,000,000kg)(57,000,000kg)] / 20m^2
This is how you got roughly 540N

So plug all your numbers in again, except change the r value to 60m^2

F = [(6.6726 x 10^-11)(57,000,000kg)(57,000,000kg)] / 60m^2

F = 60.2N

Answer 4

F=Gm1*m2/r^2

force is inversely proportional to square of distance between the centers

if distance is doubled then force will be 1/4th of the previous

so F= 540/4=135

Answer 5