Newton's Law of Gravitation Question.

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The space shuttle orbits the Earth at a height of 4.0*10^6m above the surface.
The kinetic energy of the shuttle in this orbit is 2.8*10^12J. Calculate:

a) The force of gravity on the shuttle.
b) The gravitational field strength at this height.
c) The mass of the space shuttle.

I was the thinking the first part was to do with work done? But apparently its not?

Also thinking the answer in my book could be wrong.

If anyone could help me out with the first part, I can carry on for there, or tell me the necessary formulae.

Dman

For part a) we'll assume the shuttle is undergoing a circular orbit around the earth.

In this case, the net force on the shuttle (centripetal force) is equal to the force of gravity, make sure you understand why.

Thus:

$F_{net} = F_{grav} = \dfrac{mv^2}{r}$ where $r$ denotes the distance to the centre of the earth.

However, luckily you are given it's kinetic energy $\dfrac{1}{2}mv^2 = 2.8\times 10^{12} J$.

You can manipulate the value above to give you $mv^2$. You can easily calculate the value of $r$ using the height of orbit above the surface of the earth, and the radius of the earth.

Part b) is also fairly simple!

Using the fact that $F=mg$ and Newton's Law of Gravitation, we can combine these two equations to yield:

$g = -G\dfrac{M}{r^2}$, where $M$ denotes the mass of the earth. In this case however, we are only interested in its magnitude, hence the positive value found using; $g = G\dfrac{M}{r^2}$.

Finally, for part c):

You found in part a) the force of gravity on the shuttle. Newton's Law of Gravitation again! (I'll be ignoring the negative sign in the law, as again, we are purely looking at the magnitude).

$F_{grav} = G\dfrac{Mm}{r^2}$, where m is the mass of the shuttle, M is the mass of the earth and r again is the distance to the centre of the earth. You can trivially calculate the mass of the shuttle from here.