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I find this question hard because I've only done problems involving acceleration as a function of time, and not as a function of distance. Can anyone show me how to derive an equation of acceleration as a function of time?

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I find this question hard because I've only done problems involving acceleration as a function of time, and not as a function of distance. Can anyone show me how to derive an equation of acceleration as a function of time?

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siddharth

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recon said:

I find this question hard because I've only done problems involving acceleration as a function of time, and not as a function of distance. Can anyone show me how to derive an equation of acceleration as a function of time?

You can use [tex] \frac{dv}{dt} = (\frac{dv}{dr})(\frac{dr}{dt}) = v \frac{dv}{dr} [/tex]

So I think from this you can first find v(r) and then r(t) and use that to find x.

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I know [tex]\frac{dv}{dt} = \frac{GM}{r^2}[/tex] but I can't figure out how this may be used to obtain v as a function of r.siddharth said:You can use [tex] \frac{dv}{dt} = (\frac{dv}{dr})(\frac{dr}{dt}) = v \frac{dv}{dr} [/tex]

So I think from this you can first find v(r) and then r(t) and use that to find x.

How is it possible to solve the equation [tex]\frac{d^{2}r}{dt^2}=\frac{GM}{(R+r)^2}[/tex], where R is the radius of the Earth? I have not had much calculus yet, so I have not touched on solving differential equations of this difficulty.

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siddharth

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[tex]\frac{dv}{dt} = \frac{GM}{r^2}[/tex]

Now, as I said in my previous post,

[tex] \frac{dv}{dt} = (\frac{dv}{dr})(\frac{dr}{dt}) = v \frac{dv}{dr} [/tex]

So,

[tex]v\frac{dv}{dr} = \frac{GM}{r^2}[/tex]

Now, can you find v(r) from this diff equation?

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