Object falling from 500 km altitude: finding fall time with gravitational potential energy

[gaborfk]

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 $$R_{E}\leq r\leq R_{E}+h$$, is given by

$$v=\sqrt{2GM_{E}(\frac{1}{r}-\frac{1}{R_{E}+h})}$$

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

$$\Delta t=\int^{f}_{i} dt=\int^{f}_{i} -\frac{dr}{v}$$

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

$$v= -\frac{dr}{dt}$$. 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.

 

[dextercioby]

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 $$R_{E}\leq r\leq R_{E}+h$$, is given by

$$v=\sqrt{2GM_{E}(\frac{1}{r}-\frac{1}{R_{E}+h})}$$

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

$$\Delta t=\int^{f}_{i} dt=\int^{f}_{i} -\frac{dr}{v}$$

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

$$v= -\frac{dr}{dt}$$. 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.

The first part results simply from the law of conservation of total energy of the particle's interaction with Earth.
By numerically u mean applying different methods of numerically solving integrals,e.g.the Runge-Kutta??I'm sorry,this part of computional mathematics is beyond my knowledge.
Hopefully somebody else would guide through point b).

Daniel.

 

[HallsofIvy]

The force law is F= -GmM/r² so the potential energy at r is the integral of Fx dx from R_E to r: GMm(1/r- 1/R_E). Of course, the potential energy at R_E+ h is given by the same formula: GMm(1/(R_E+h)- 1/R_E). The change in potential energy is the difference of those:
GMm(1/r- 1/(R_E+h)) (the "1/R_E" term cancel) and that must be equal to the acquired kinetic energy:
(1/2)mv²= GMm(1/r- 1/(R_E+h)) . Solve for v.

Now, you want to integrate $$\int^{f}_{i} -\frac{dr}{v}$$
which, from the previous part is
$$-\frac{1}{\sqrt{2GM}}\int^{R_E}_{R_E+h}\frac{dr}{\sqrt{1/r-1/(R_E+h)}}$$

If you are a computer science major, you certainly ought to be able to integrate that numerically, using, say, Simpson's rule. Any calculus text will describe it.

(dextercioby: "Runge-Kutta" is a numerical technique for differential equations. Not necessary here.)