Question

The average distance separating earth and the moon is 384,00kg (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

given

distance between The Earth and the moon, d = 384000 km

mass of the satellite, m = 2.75*10^4 kg
we know,
M_earth = 5.97*10^24 kg
M_moon = 7.35*10^22 kg
r = d/2 = 384000/2 = 192000 km = 1.92*10^8 m

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

Fnet = F_earth - F_moon

= G*M_earth*m/r^2 - G*M_mmon*m/r^2

= G*(M_earth - M_mmon)*m/r^2

= 6.67*10^-11*(5.97*10^24 - 7.35*10^22)*2.75*10^4/(1.92*10^8)^2

= 293.4 N <<<<<<<------------Answer

b) Let v is the velocity of the satellite at the given point.

Apply conservation of energy

Uf + KEf = Ui + KEi

-G*M_earth*m/(Re + h) - G*M_moon*m/(d - Re - h) + (1/2)*m*v^2 = -G*M_earth*m/r - G*M_moon*m/r

-6.67*10^-11*5.97*10^24*2.75*10^4/(6400*10^3 + 5500*10^3) - 6.67*10^-11*7.35*10^22*2.75*10^4/(1.92*10^8 - 6400*10^3 - 5500*10^3) + (1/2)*2.75*10^4*v^2 = -6.67*10^-11*5.97*10^24*2.75*10^4/(1.92*10^8) - 6.67*10^-11*7.35*10^22*2.75*10^4/(1.92*10^8)

-9.209*10^11 + (1/2)*2.75*10^4*v^2 = -5.773*10^10

v = 7923 m/s <<<<<<<<<--------------------------Answer