(adsbygoogle = window.adsbygoogle || []).push({}); An object is released from rest at an altitude h above the surface of the Earth.

a) Show that its speed at a distance r from Earth's center, where [tex]R_{E}\leq r\leq R_{E}+h[/tex], is given by

[tex]v=\sqrt{2GM_{E}(\frac{1}{r}-\frac{1}{R_{E}+h})}[/tex]

b)Assume the release altitude is 500km. Perform the integral

[tex]\Delta t=\int^{f}_{i} dt=\int^{f}_{i} -\frac{dr}{v}[/tex]

to find the time of fall as the object moves from the release point to the Earth's surface. The negative sign appears because the object is moving opposite to the radial direction, so its speed is

[tex]v= -\frac{dr}{dt}[/tex]. Perform the integral numerically.

I am at a total loss with this one. It is my final physics problem ever! I am done with physics! (Computer Science major) Please, help. My head is about to explode.

Thank you.

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