Question on Elliptic Orbits. Difficult.

[Wesc]

A satellite undergoes an elliptic orbit about the Earth of mass M, with maximum
distance 6R and minimum distance 3R from the Earth's centre.

(a) Show that twice the minimum velocity v(min) = the maximum velocity v(max) = (2/3)*sqrt(GM/R)

(b) Show eccentricity = 1/3

We are told that we can assume the orbit is described by 1/r = (1 + e*cos(Theta))/L where where r is the distance from the Earth’s centre, e is the eccentricity with 0 ≤ e < 1
and l = h^2/GM for constant h, the angular momentum per unit mass

How I started off was using the conservation of angular momentum and from that got v(max) = 2*v(min) . Tried conservation of energy but got nowhere there. Also tried other equations but I'm not getting anywhere! If anyone could show me a solution I would be so grateful :) Thank you.

 

[voko]

Can you show exactly how conservation of energy got you nowhere?

For (b), what are the min/max values of the RHS of the equation of motion you were given?

 

[Wesc]

Can you show exactly how conservation of energy got you nowhere?

For (b), what are the min/max values of the RHS of the equation of motion you were given?

Hey, thanks for the feedback.

Well for conservation of energy I did this:

E = (1/2)*m*v(max)^2 - G*M*m/r(min) = (1/2)*m*v(min)^2 - G*M*m/r(max)
I then canceled the 'm's .. and I'm not sure if you can really extract any other information from there??

And if you're talking about this equation of motion 1/r = (1 + e*cos(Theta))/L then the max is at cos(Theta) = 1 and min at cos(theta) = -1 but I'm just not sure what the next step is? I'm just a bit lost. :/

 

[voko]

Hey, thanks for the feedback.

Well for conservation of energy I did this:

E = (1/2)*m*v(max)^2 - G*M*m/r(min) = (1/2)*m*v(min)^2 - G*M*m/r(max)
I then canceled the 'm's .. and I'm not sure if you can really extract any other information from there??

You have obtained the relationship between Vmax and Vmin, and you were given the relationship between Rmax and Rmin. Use them.

And if you're talking about this equation of motion 1/r = (1 + e*cos(Theta))/L then the max is at cos(Theta) = 1 and min at cos(theta) = -1 but I'm just not sure what the next step is? I'm just a bit lost. :/

That is correct, plug that into the equation and use Rmax and Rmin.

 

[Nickie]

I would like to first commend the questioner for their efforts in trying to solve this problem. It can be challenging to tackle complex equations and concepts in physics, but perseverance and seeking help is key to understanding and solving difficult problems.

Now, let's break down the problem and go through the steps to solve it.

(a) To show that v(min) = v(max) = (2/3)*sqrt(GM/R), we need to use the conservation of energy and angular momentum.

Conservation of energy tells us that the total energy of a satellite in orbit is constant, and it is given by the sum of its kinetic and potential energies. Mathematically, it can be written as:

E = 1/2 * m * v^2 - G * M * m / r

Where E is the total energy, m is the mass of the satellite, v is its velocity, G is the gravitational constant, M is the mass of the Earth, and r is the distance between the satellite and the Earth's center.

Since we are dealing with an elliptic orbit, the satellite's distance from the Earth's center will vary between the minimum distance (r(min)) and the maximum distance (r(max)). So, we can rewrite the equation as:

E = 1/2 * m * v(min)^2 - G * M * m / r(min) = 1/2 * m * v(max)^2 - G * M * m / r(max)

We can then rearrange the equation to solve for v(min) and v(max):

v(min) = sqrt(2 * G * M * (1/r(min) - 1/(2*r(max))))

v(max) = sqrt(2 * G * M * (1/r(max) - 1/(2*r(min))))

Now, we are given that the maximum distance is 6R and the minimum distance is 3R, so we can substitute those values into the equations:

v(min) = sqrt(2 * G * M * (1/3R - 1/(2*6R))) = sqrt(G * M / 3R)

v(max) = sqrt(2 * G * M * (1/6R - 1/(2*3R))) = sqrt(2 * G * M / 3R)

Since we have v(min) = sqrt(G * M / 3R) and v(max) =