At what distance from the Earth center (in units of km) will an object of mass m weigh the same as it does on?

Question

At what distance from the Earth center (in units of km) will an object of mass m weigh the same as it does on the surface of the Moon?

Answer 1

The basic solution is based on Newton's famous gravitational formula F=gmM/r*r where g is the universal gravitational constant, m and M are the two different masses involved and r is the distance between their mass centers of gravity. For spheres, this is the center of the sphere. This formula calculate the force experienced my an object of mass m when it is a distance r from another object of mass M.



To "brute force" this calculation, you would need to know the mass of the moon and the radius of the moon (distance to the surface from the center of the mass M). You could then calculate the force on an object on the surface of the moon.



Then, replace M with the mass of the earth and solve for the distance from the earth's center of mass substituting in the force you calculated on the moon.



Now while this is the root of the answer, there are certainly other ways to actually calculate it as explained in some of the other answers. For example, you could look up the gravitational force that would be experienced by an object of mass m on the surface of the moon and you wouldn't need to know the mass of the moon (even though it is still implied).

Answer 2

There are two solutions to this.



1. below the surface. As you move down below the surface towards the center of the earth, gravity decreases, hitting zero at the center. At some point it will equal the moon, which is about 1/6 of the earth's gravity. To calculate this is complicated due to variations in density with depth. It's about 100 km from the center from the references.



2. Above the surface, gravity falls off via the inverse square law

earth, variation of g with height

g = GM/r²

g is acceleration from earth attraction in m/s²

earth radius 6,371 km = 6.37e6 meters

earth mass M 5.97e24 kg

earth GM = 3.98e14

moon g 1.622 m/s²

g = (3.98e14) / r²

r² = 3.98e14 / 1.622

r = 1.57e7 meter

h = 6.37e6 + 1.57e7 = 2.20e7 m

or 2.20 e4 km

or 22000 km

Answer 3

Gravity of moon = 1/6 of gravity at earth's surface (or maybe you were given a more exact number. I think it might be 0.17 but I'm not sure off the top of my head).



F is proportional to 1/r^2.



So r^2 is proportional to 1/F, or r is proportional to 1/sqrt(F). Set up the ratio, with r/(r at surface of earth).



Edit: Surface gravitational acceleration of the moon is 1.62 m/s^2. That's 1.62/9.8 = 0.165 of earth's gravity.

Answer 4

W1 = W2

m.g1 = m.g2

g1 = g2

G.M(moon) / r^2 = G.M(earth) / x^2

M(moon) / r^2 = M(earth) / x^2

x^2 = (M(earth).r^2) / M(moon)

That's it, u just have to insert the number



W1= weight on moon

W2= weight x distance from earth center

m= object mass

g1= gravity on moon

g2= gravity x distance from earth center

M(moon)= moon mass

r= moon radius

M(earth)= earth mass

x= the distance from earth center

Answer 5

That is quite difficult to calculate.



The Earth has a massive metal core so as a person gets closer to it, his weight increases. Then as he continues to the center, his weight decreases.