Question

In deep space, sphere A of mass 94 kg is located at the origin of an x axis and sphere B of mass 100 kg is located on the axis at x = 1.8 m. Sphere B is released from rest while sphere A is held at the origin. (a) What is the gravitational potential energy of the two-sphere system just as B is released? (b) What is the kinetic energy of B when it has moved 0.60 m toward A?

Answer 1

(a) $-3.48\cdot 10^{-7} J$

The gravitational potential energy of the two-sphere system is given by

$U=-\frac{Gm_A m_B}{r}$ (1)

where

G is the gravitational constant

$m_A = 94 kg$ is the mass of sphere A

$m_B = 100 kg$ is the mass of sphere B

r = 1.8 m is the distance between the two spheres

Substitutign data in the formula, we find

$U=-\frac{(6.67\cdot 10^{-11})(94 kg)(100 kg)}{1.8 m}=-3.48\cdot 10^{-7} J$

and the sign is negative since gravity is an attractive force.

(b) $1.74\cdot 10^{-7}J$

According to the law of conservation of energy, the kinetic energy gained by sphere B will be equal to the change in gravitational potential energy of the system:

$K_f = U_i - U_f$ (2)

where

$U_i=-3.48\cdot 10^{-7} J$ is the initial potential energy

The final potential energy can be found by substituting

r = 1.80 m -0.60 m=1.20 m

inside the equation (1):

U=-\frac{(6.67\cdot 10^{-11})(94 kg)(100 kg)}{1.2 m}=-5.22\cdot 10^{-7} J

So now we can use eq.(2) to find the kinetic energy of sphere B:

$K_f = -3.48\cdot 10^{-7}J-(-5.22\cdot 10^{-7} J)=1.74\cdot 10^{-7}J$