physics question?

Question

What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/34 of its value at the Earth's surface?

Answer 1

We know that
F=GMm/R^2
let r be the distance where acceleration a due to gravity is 1/34 g.
so

At the surface F=mg where R=radius of the Earth [from its center to the equator and equals 6,378.135 km see ref]
Some distance away r it is F2=m(1/34) g

F/F2=mg/[m(1/34) g]=GMm/R^2/GMm/r^2
34=(r/R)^2

Finally

r=Rsqrt(34)

Have fun!

Answer 2

It's been a while for me to talk about physics but I think you need to know the mass of the object that the Earth is pulling on before you can measure the gravitational effects on it. The gravitational law is F= G (m1m2)/r(2) where:

F is the magnitude of the gravitational force between the two point masses,
G is the gravitational constant (6.67 × 10−11 N m2 kg−2)
m1 is the mass of the first point mass,
m2 is the mass of the second point mass,
r is the distance between the two point masses.

Like I said, it's been a while but I don't think you can answer this with out the mass of the second body (the first body being Earth).

Cheers