Question

Two spheres of masses m = 1.08 g and m' = 1.09 102 kg are isolated from all other bodies and are initially at rest, with their centers a distance r = 16 cm apart. One minute later, the smaller sphere has moved 0.570 mm toward the larger sphere.

Compute the acceleration

and the value of G.

Answer 1

Given data
Mass of the spheres,
m = 1.08 g
= (1.08 g) (10^-3 kg)/ (1 g)
= 1.08 * 10^-3 kg
m' = 1.09 *10^2 kg
Initial distance between m and m' is, r = 16 cm
= (16 cm )(10-2 m)/ (1 cm)
= 0.16 m
Initial velocities of the two spheres are zero.
After time interval, t = 60 s, the distance moved by m is,
d = 0.570 mm
= (0.570 mm)(10^-3 m)/ (1 mm)
= 0.570 *10^-3 m

The acceleration due to gravity is calculated by using the
equations of motion, that is,
d = vo t + 1/2 g t2
0.570 *10^-3 m = 0 + 1/2*g* (60 s)2
g = 3.17 *10^-7 m/s2
_________________________________________________

The realtion between acceleration due to gravity and gravitational
constant is,
g = Gm' / r2
(3.17* 10^-7 m/s2) = G (1.09 *10^2 kg ) / (0.16m )2
G = 7.40 * 10-11 N.m2 / kg2