Elliptical Orbit Question with no numbers

In summary, the conversation discusses the physics of a satellite in an elliptical orbit around the Earth and explores various equations and concepts related to its motion. The equations used include the force of gravity, gravitational potential energy, total energy, angular momentum, and velocity at different points along the orbit. The conversation also mentions the work-energy principle and the use of integrals to calculate work. The final questions involve finding the speed and work needed to change the satellite's orbit from elliptical to circular.
  • #1
vasilicus
6
0

Homework Statement



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.


Homework Equations



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


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.
 
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  • #2
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
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
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
Bump still stuck
 
  • #6
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
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
You are supposed to set up an integral. What integral would you use to calculate work?
 
  • #9
Um,
W = integral(xi-xf)(Fxdx)?
 
  • #10
More generally,

[tex]\int_{x_i}^{x_f}{\boldsymbol F}\cdot d{\boldsymbol l}[/tex]

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.)
 

FAQ: Elliptical Orbit Question with no numbers

1. What is an elliptical orbit?

An elliptical orbit is a type of orbit in which an object revolves around another object in an oval or elliptical shape, rather than a perfect circle. This is due to the gravitational pull of the two objects and the shape of the orbit is determined by their masses and distance from each other.

2. How is an elliptical orbit different from a circular orbit?

An elliptical orbit differs from a circular orbit in the shape of the path that the object takes. In a circular orbit, the object follows a perfect circle around another object, while in an elliptical orbit, the path is in the shape of an oval or ellipse.

3. What causes an object to have an elliptical orbit?

An object has an elliptical orbit due to the gravitational pull of the two objects involved. The shape of the orbit is determined by the mass and distance of the two objects. The larger the mass and the closer the distance, the more circular the orbit will be. As the mass and distance change, the shape of the orbit becomes more elliptical.

4. Can an object have an elliptical orbit around more than one object?

Yes, an object can have an elliptical orbit around more than one object. This is known as a binary orbit, where the object is orbiting around two objects simultaneously. This can occur in systems with multiple stars or planets.

5. How is the speed of an object in an elliptical orbit calculated?

The speed of an object in an elliptical orbit is calculated using Kepler's third law, which states that the square of the orbital period is proportional to the cube of the semi-major axis. This means that the speed of the object will vary depending on its distance from the other object and the time it takes to complete one orbit.

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