How to Calculate Gravitational Field Strength at 400000m from Earth?

[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?

 

[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)!

 

[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.