The gravitational field strength is also governed by an inverse square law. What is the gravitational field strength 200km above the Earth's surface, at the altitude of many manned space flights?

Question

beth12345 · Answer

The answer I'm given is 9.2N/kg but U don't know how to get it. I also have an equation sheet with the following equations...
$$F _{g}=m \times g$$$$F=G \times ((m _{1} \times m _{2})/r ^{2})$$$$F _{fr}= \mu \times F _{N}$$$$F _{net}= m \times a$$$$p=m \times v$$$$\Delta p= F \times \Delta t$$$$F=k \times x$$
 but I also don't know what one to use, if someone could help me, that would be nice :)

beth12345 · Answer

I* don't know how to get it

apoorvk · Answer

okay in the formula for gravitation force just add to 200km to the earth's radius of 6400kms... and calculate... although there is a more generic formula http://en.wikipedia.org/wiki/Gravity_of_Earth#Altitude

beth12345 · Answer

so would I use the second equation?

beth12345 · Answer

or the first?

apoorvk · Answer

you mean on the wiki page? ofcourse the first because you need to find it for above the surface, at an 'altitude'

apoorvk · Answer

in second equation if you need it for future questions, take rho nought and rho 1 as same, so the cancel out.

beth12345 · Answer

so is the equation $$g _{h}=g _{0}\left[\begin{matrix}r e /\ re + h\end{matrix}\right]^{2}$$?