Question

The average distance separating earth and the moon is 384,000 kg (center to center). What is the Net gravitational force exerted by the earth and the moon on a 2.75x10⁴ kg spaceship located half way between them (take the direction toward earth to be positive? if the spaceship falls toward earth starting from rest at a point where it is located 55,00km above the surface of the earth, what is its velocity at that point?

Answer 1

(a) Average distance between earth and moon = 384000 km = 384000 x 10^3 m

So, mid-point of this distance, R = (384000 x 10^3 m) / 2 = 192000 x 10^3 m

Mass of earth, m1 = 5.972 x 10^24 kg

Mass of moon, m2 = 7.348 x 10^22 kg

Mass of the spaceship, m = 2.75 x 10^4 kg

Net gravitational force exerted by earth and moon on the spaceship,

     F = G*m1*m / R^2 – G*m2*m / R^2

=> F = G*m(m1 – m2)/R^2

Put the values –

F = (6.674 x 10^-11 x 2.75 x 10^4)*( 5.972 x 10^24 - 7.348 x 10^22) / (384000 x 10^3)^2

   = 7.342 x 10-10 x 10^11 = 73.42 N

(b) Here we will make an assumption. Since the gravitational pull due to moon at such a large distance is small so it can be neglected.

So, potential energy lost by the spaceship = - G*m1*m/R+ G*m1*m/R’

                                                                            = G*m1*m[1/R’ – 1/R]

This loss in energy will be equal to the gain in the kinetic energy.

=> G*m1*m[1/R’ – 1/R] = (1/2)*m*v^2

=> G*m1*[1/R’ – 1/R] = (1/2)*v^2

=> v^2 = 2*G*m1*[1/R’ – 1/R] = (2*6.674 x 10^-11*5.972 x 10^24)*(1/5500 – 1/192000)*10^-3

             = 0.01408 x 10^10

=> v = 0.1187 x 10^5 m/s = 11.87 km/s