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ok guys, i've got this question here and although I know it shouldn't be difficult i just can't work it out;

there is a point on the line joining the centres of the earth and the moon where the gravitational field strength is zero how far is this from the centre of the earth?

M of moon=7.4 x 10²² kg

M of earth=6.0 x 10 >24 kg

dist between moon and earth =3.8 x 10 to 8 metres

radius of earth=6.4 x 10 to 6

radius of moon = 1.7 x 10 to 6

so far I have ; g= -GM/r²

into; M moon/(r-x)² = M earth/ x²

but just can't get any further and don't even know if this is right, i feel like such a dumbass. Can anyone help please?

I've tried using ratios and i kniow what the answer is as it's in the back of book, but not sure how to get it!

there is a point on the line joining the centres of the earth and the moon where the gravitational field strength is zero how far is this from the centre of the earth?

M of moon=7.4 x 10²² kg

M of earth=6.0 x 10 >24 kg

dist between moon and earth =3.8 x 10 to 8 metres

radius of earth=6.4 x 10 to 6

radius of moon = 1.7 x 10 to 6

so far I have ; g= -GM/r²

into; M moon/(r-x)² = M earth/ x²

but just can't get any further and don't even know if this is right, i feel like such a dumbass. Can anyone help please?

I've tried using ratios and i kniow what the answer is as it's in the back of book, but not sure how to get it!

$M_{m}$ is the Moon's mass. $M_{e}$ is the Earth's mass.

You've derived

$\frac{M_{m}}{(r-x)^{2}} = \frac{M_{e}}{x^{2}}$

$x^{2}M_{m} = (r-x)^{2}M_{e}$

$x^{2}M_{m} = (x^{2}-2xr+r^{2})M_{e}$

$x^{2}\frac{M_{m}}{M_{e}} = x^{2}-2xr+r^{2}$

$x^{2}(\frac{M_{m}}{M_{e}}-1) + 2rx - r^{2} = 0$

You can now apply the quadratic formula to this since you know the values of the Earth and Moon's mass and r. One result will be positive, the other negative, you want the positive one.

You've derived

$\frac{M_{m}}{(r-x)^{2}} = \frac{M_{e}}{x^{2}}$

$x^{2}M_{m} = (r-x)^{2}M_{e}$

$x^{2}M_{m} = (x^{2}-2xr+r^{2})M_{e}$

$x^{2}\frac{M_{m}}{M_{e}} = x^{2}-2xr+r^{2}$

$x^{2}(\frac{M_{m}}{M_{e}}-1) + 2rx - r^{2} = 0$

You can now apply the quadratic formula to this since you know the values of the Earth and Moon's mass and r. One result will be positive, the other negative, you want the positive one.

Dude no I get two positive answers ffs

Original post by Confusedguyskrrt

Dude no I get two positive answers ffs

I wouldn't bother with the quadratic stuff.

M1 = earth mass, M2 = moon mass

x1 = dist (g = 0 to earth), x2 = dist (g = 0 to moon)

At g = 0: (GM1)/(x1)^2 = (GM2)/(x2)^2

=> (x1/x2)^2 = M1/M2

=> x1/x2 = Sqrt[(6 x 10^24)/(7.4 x 10^22)] = 9

Point lies 90% of seperation distance from earth

=> 0.9 x (3.8 x 10^8) = 3.42 x 10^8 m.

I assumed seperation distance given refers to between centres. If between surfaces, add radii first.

(edited 4 years ago)

I know you posted this ages ago but I just found it and it really helped me so thanks!

Original post by Physics Enemy

I wouldn't bother with the quadratic stuff.

M1 = earth mass, M2 = moon mass, x1 = dist g = 0 to earth, x2 = dist g = 0 to moon

At g = 0: (GM1)/(x1)^2 = (GM2)/(x2)^2

=> (x1/x2)^2 = M1/M2

=> x1/x2 = Sqrt[(6 x 10^24)/(7.4 x 10^22)] = 9

Point lies 90% of seperation distance from earth

=> 0.9 x (3.8 x 10^8) = 3.42 x 10^8 m.

I assumed seperation distance given refers to between centres. If between surfaces, add radii first.

M1 = earth mass, M2 = moon mass, x1 = dist g = 0 to earth, x2 = dist g = 0 to moon

At g = 0: (GM1)/(x1)^2 = (GM2)/(x2)^2

=> (x1/x2)^2 = M1/M2

=> x1/x2 = Sqrt[(6 x 10^24)/(7.4 x 10^22)] = 9

Point lies 90% of seperation distance from earth

=> 0.9 x (3.8 x 10^8) = 3.42 x 10^8 m.

I assumed seperation distance given refers to between centres. If between surfaces, add radii first.

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