When a spacecraft travels from the Earth to the Moon, both the Earth and the Moo

Question

When a spacecraft travels from the Earth to the Moon, both the Earth and the Moon exert a gravitational force on the spacecraft. Eventually, the spacecraft reaches a point where the Moon's gravitational attraction overcomes the Earth's gravity. How far from the Earth's surface must the spacecraft be for the gravitational forces from the Moon and the Earth to just cancel?

_______m

You are an astronaut

(m = 100 kg)

and travel to a planet that is the same radius and mass as the Earth, but it has a rotational period of only 8 h. What is your apparent weight at the equator of this planet?
_______ N

Explanation / Answer

That is likely a good answer, except the canceled gravity point is shifted significantly by the gravity of the Sun. On the average the distance is about 77,000 kilometers (if I remember correctly) above the surface of the Moon at the center of the side of the Moon which faces Earth. At all altitudes above Earth's surface, only partial cancellation occurs, except in approximately a straight line between the Moon and Earth.

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write Newton's second law for an object sitting on the equator of this planet

sum of forces = ma

the forces acting on this object are the normal force (acting up), weight (acting down), and these combine to produce a centripetal force (acts down)

setting down as the negative direction, we have:

N - mg = -mv^2/r

since we are told that the planet has the same mass and radius as the earth, we know the value of g on this planet will be the same as on the earth

then we have for the normal force (what you would appear to weight if you stood on a scale):

N=m(g-v^2/r)

the velocity of a point on the equator is 2 pi r/8 hours,
so the term v^2/r becomes:

(2 pi r)^2/(r x 8 hrs) = 4 pi^2 r/8x3600s

v^2/r = 8.77 m/s/s using r = 6.4x10^6 m

so the term g-v^2/r = 9.8-8.77=1.03 m/s/s

and the apparent weight is 100 kg x 1.03 m/s/s=103 N

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