# Deduce gravitational strength

1. Mar 24, 2006

### Harmony

Distance form the earth's surface: Gravitational potential/MJkg-1
0 : -62.72
400000 : -59.12

Deduce the earth's gravitational field strength at a height of 400000m.

Since g=-GM/r^2 and V=-GM/r , I tried to solve this question using V/r. But I failed to get the correct answer. Any suggestion on how to approach the question?

Last edited: Mar 24, 2006
2. Mar 24, 2006

### Tide

I don't understand your notation in the first paragraph.

However, be sure to use the distance from the center of the Earth in your calculation (and NOT the height above the surface)!

3. Mar 24, 2006

### andrevdh

The gravitational field is given by
$$F_G/m=-\frac{GM}{r^2}$$
so at the surface it is $-9.8$
and at the required distance $-8.4$ mega newtons per kilogram.

Last edited: Mar 24, 2006