how to get the distance from earth when the net gravitational force of the spaceship is zero?

Question

sorry for the pressure. but can someone answer this problem? actually i don't know what solution to use.

this is the situation: a spaceship is on a straight line path between the earth and its moon. at what distance from earth is the net gravitational force of the spaceship is zero?

it will really help a lot. thanks for your concern.

Answer 1

Since the 1st answer's equation can't be read, I'll go through it also.
Me = Earth mass, Mm = moon mass, xt = distance between Earth and moon, x = zero-g distance from Earth
GMe/x^2 = GMm/(xt-x)^2
x = xt/sqrt(Mm/Me)+1)
Mm/Me = 0.0123 (ref.)
xt = 3.844E8 m (ref.)
x = 3.460E8 m (answer)

Answer 2

you'll be able to desire to locate the area the forces are the comparable. the terrific equation for the strain of gravity is F=M*m*G/d^2 moon's mass= 7.3x10^22 kg (i will label this as Mmoon) Earth's mass=5.9742 × 10^24 kg. (Mearth) finished distance between the earth and moon (d)=384,403 Km. Gearth=9.80 2 m/s^2 Gmoon=a million.6 m/s^2 m=spaceships mass (which will cancel out, so do now not project approximately it) finished distance is d, and we are going to label the hollow from earth to spaceship as x. (meaning the hollow from the moon to the spaceship is d-x) So whilst Fmoon=Fearth we've Mearth*m*Gearth/x^2=Mmoon*m*Gmoon/(d-x..... like I suggested till at last now m cancels from the two aspects so we've Mearth *Gearth/x^2=Mmoon*Gmoon/(d-x)^2 now therapy for x. in basic terms be sure you exchange gadgets to be the comparable (ie convert km to m) I also have a feeling that it truly is a physics homework activity and that i could desire to journey undesirable as a instructor to in uncomplicated words supply the respond, yet I helped set it up for you so which you in uncomplicated words might desire to "plug and chug"

Answer 3

gravitaional force form the earth or the moon is = GxM/r^2

5.98x10^24/(X)^2=7.35x10^22/(3.84x10^8-X)^2

ans=3.457x10^8m