Homework Help: Elliptical Orbit Question with no numbers

1. Jan 18, 2009

vasilicus

1. The problem statement, all variables and given/known data

A satellite of mass m is in an elliptical orbit around the Earth, which has mass M and radius R. The orbit varies from closest approach of 'a' at point A to maximum distance of 'b' from the center of Earth at point B. At point a the speed of the satellite is vo. Assume Ug = 0 when masses are an infinite distance apart. Express your answers in terms of vo, a, b, m, M, R, and G.

A. Write the definite integral (including limits) that can be evaluated to show that Ug at distance r from the center of the Earth is given by Ug = -GMm/r

B. Determine the total energy at A.

C. Determine the angular momentum at A.

D. Determine the velocity at B.

As the satellite passes point a, it changes to a circular orbit of radius a around the center.

E. Determine the speed.
F. Determine the work done by a rocket engine to make this change happen.

2. Relevant equations

Fg = GMm/r
f = GM/r2
g = Fg/m
U = -GMm/r
E = K + U = 1/2mv2 - GMm/r

3. The attempt at a solution

For a. I wasn't sure where to start. What am I supposed to do, reverse-derive the equation? If so could someone show me how because I really don't know.

b should just be E = mvo2/2 - GMm/r but I'm not sure this seems to easy.
c like b I already know that the equation is mvoa but again it seems to easy that it's just asking to repeat an equation I learned in class, there must be a catch.
d I thought v1r1 = v2r2 therefore vb = voa/b would work but my teacher said this was the answer:
vb = square root of (vo2 + 2GM(1/b - 1/a)) I plugged in numbers and the % difference between the equations was approximately 1.8%. How did he get that equation? Why isn't mine right?

e my book says that v = square root of (GM/r) so I put square root (GM/a) but again it just seems way too easy.

f not sure where to start.

2. Jan 18, 2009

vasilicus

Please help somebody, the test is Tuesday and a similar problem will be worth 1/3rd of the points on the test and I still don't know how to do it...

3. Jan 18, 2009

D H

Staff Emeritus
a. How are force and potential related?

b. Correct -- if the 'a' is the distance from the center of the Earth rather than the distance to the surface of the Earth, that is. E=1/2mv2-GMm/r is valid at any point on the orbit and is constant throughout the orbit.

c. Correct (but same caveat regarding radial distance versus altitude applies). The simple expression you used is valid only at perigee and apogee. Why?

d. v1r1 = v2r2 is incorrect: It doesn't conserve energy. You have the wrong equation. Check your notes, your text, or google "vis-viva equation".

e. Use the work-energy principle.

4. Jan 18, 2009

vasilicus

for a.

Ug = mgy
g = Fg/m
Fg = GMm/r2
therefore g = GM/r2
therefore Ug = GMmy/r2
y = r therefore Ug = GMM/r but U can't be positive so it's -GMm/r

But how do you write the appropriate definite integral?

For e. and f. Could you please elaborate, I looked again and tried to look up some stuff but I'm still stuck...

5. Jan 19, 2009

vasilicus

Bump still stuck

6. Jan 19, 2009

D H

Staff Emeritus
What you did in post #4 is invalid. Ug=mgy is valid only near the surface of the Earth.

What is the definition of potential energy? (Hint: Work).

7. Jan 19, 2009

vasilicus

Well I know work is the change in Ug and the Gravitational Potential Energy of two particles separated by a distance r is -GMm/r

Would e. just be GM/a? This seems to easy but then so did b. and c.
And then for f if Work = Change Ug then would it be
GMm/rf - GMm/ri
GMm/a - 2GMm/(a+b) = Work?

8. Jan 19, 2009

D H

Staff Emeritus
You are supposed to set up an integral. What integral would you use to calculate work?

9. Jan 19, 2009

vasilicus

Um,
W = integral(xi-xf)(Fxdx)?

10. Jan 19, 2009

D H

Staff Emeritus
More generally,

$$\int_{x_i}^{x_f}{\boldsymbol F}\cdot d{\boldsymbol l}$$

So, to compute gravitational potential at some distancer from a point mass, what is the force? What are the limits of integration? (Hint: gravitational potential goes to zero as distance tends to infinity.)