How Do You Derive the Radius Equation for a Satellite's Elliptic Orbit?

[Cuttlas]

Homework Statement

Our teacher gave us a formula and ask us to proof it, when I asked him for more detail he told I'm looking for radius equation in this formula.
The formula is:

G.Me.m/r*2=m.d*2r/dt*2

. means Multiply
/ means Divide
* means Power, r to the power of 2 is r*2

I believe target section is ---> d*2r/dt*2

Me is the mass of earth, G gravity, m is the mass of satellite.

Homework Equations

Nothing Special

The Attempt at a Solution

, This question is related to Elliptic motion of a Satellite
and Just want to say this question is urgent and I have only 2 or 3 days left.

***Excuse me for my bad English and also excuse me if I'm doing something wrong here, because I'm too newbie!

Thanks For any help Friends

 

[D H]

It is unclear what you are supposed to prove: Is it

• $G M_e m/r^2 = m\frac{d^2r}{dt^2}$, or
• That objects subject to the above will follow an elliptical path?

 

[Cuttlas]

Hi D H

Yes the formula to proof is the one you wrote, and yes this formula is related to a satellite in a elliptical path. so is there anything unclear yet?

 

[D H]

so is there anything unclear yet?

Yes, there is something unclear. You didn't answer my question. What, exactly, are you supposed to prove?

 

[Cuttlas]

I should find this section d2r/dt2, our teacher told you should find the result of d2r/dt2 and then by the formula above find the radius equation. so?

P.S: How did you wrote that formula in the mathematics form, I can not find any tool for typing math formulas here!

 

[D H]

You still are not being clear. I see two possible interpretations of your question.

Interpretation #1. You are supposed to prove $G M_e m/r^2 = m\frac{d^2r}{dt^2}$
If this is the problem, you should not have posted this question in the advanced physics section. It is instead an introductory physics level (high school physics to freshman/sophomore calculus-based physics) problem.

Interpretation #2. You are supposed to prove that orbits are elliptical. If this is the case, this is an advanced physics problem. The proof is not simple, however, and it is typically not given as an assignment. Instead, most sophomore/junior level classical mechanics texts walk the reader through the proof over multiple pages.That you posted this in "Advanced Physics" is part of the confusion. Is this truly an advanced physics question, where you are asked to prove that orbits are elliptical, or is it an introductory physics question where you are asked to prove that simple math equation (and the elliptical nature of the orbit does not come into play)?

 

[Cuttlas]

Thanks for your attention D H, I respect it a lot.

I think it is something between 1 and 2 ! As I can remember I should only prov the formula and we know it is a elliptical orbit. so maybe introductory physics level.

But maybe the target is both, I mean proving the formula and proving that orbits are elliptical.

BTW for now I think it is better to know it as a introductory physics level problem, so how to prove the formula (we know it is a elliptical orbit)?

 

[D H]

How to prove the formula? Simple. Use Newton's universal law of gravitation and Newton's second law of motion.

 

[Snicker]

You may prove Newton's law of gravity from ALL of Kepler's three laws. I know two proofs -- a geometrical one by Newton and an analytical one by Jacque Binet. Both proofs are very difficult and require several out-of-the-way lemmas. The problem is just too difficult.

If, instead, you are to prove the satelites traverse elliptical paths, your task is tractable. Just change to polar coordinates and do some integratin'.

 

[Cuttlas]

How to prove the formula? Simple. Use Newton's universal law of gravitation and Newton's second law of motion.

But How? Can you write it please?

 

[D H]

But How? Can you write it please?

No, I can't. We don't do your homework for you. We help you do your own homework.

I gave you a huge hint. What are Newton's law of gravitation and Newton's second law of motion? Write them down.

 

[Cuttlas]

Thanks D H But...

Newton's law of gravitation=

Newton's second law of motion= F=ma

If I want to use these then I have:

G.Me.m/r*2=m.r.ω*2

But it is for circular motion not Elliptic motion. It is not a homework, It is just a project out of students knowledge and they should research for it, the one who can solve it will gain extra score, the one who can't nothing!

So I'll really appreciate you a lot if you can help me solve this problem for extra score :)

So what is (d2r/dt2) ? How to calculate this and use it in the formula?
Look at this PDF file: http://web.aeromech.usyd.edu.au/AERO2705/Orbit%20Geometries.pdf

Is it useful for solving this problem?

something more, on another forum someone gave me this formula too,

But I can't understand it :(

So what should I do now?

 

[D H]

Newton's law of gravitation=

Newton's second law of motion= F=ma

Equality is a transitive relationship. In other words, if A=B and A=C, then B=C. So use this.

What is the "a" in F=ma?

 

[Cuttlas]

Equality is a transitive relationship. In other words, if A=B and A=C, then B=C. So use this.

What is the "a" in F=ma?

Did you check that PDF?

About F=ma, a is (rω²) means acceleration but it is in circular motion, in Elliptic motion I do not know what is it??

I really do not know how to solve this problem, what is (d2r/dt2) exactly? what should do with it? your answers is really short with small details. and I'm getting mixed up !

 

[Whovian]

About F=ma, a is (rω²) means acceleration but it is in circular motion, in Elliptic motion I do not know what is it??

Ah, no. You're going to have other components of velocity that aren't parallel to the ellipse. Try thinking of acceleration as the conventional second derivative. Be very careful what you're differentiating with respect to each time, though.

 

[Cuttlas]

Ah, no. You're going to have other components of velocity that aren't parallel to the ellipse. Try thinking of acceleration as the conventional second derivative. Be very careful what you're differentiating with respect to each time, though.

My exact question is that what should I write instead of (d*2r/dt*2) in the formula?
Really what should I write?

 

[andrien]

well you have gotten the exact equation.you should use now the constancy of moment of momentum because the force is central in nature to elliminate that d(theta)/dt term.if you still have problem see any dynamics book in which a chapter on central orbit is given like A.S.RAMSEY DYNAMICS VOL. 1.

 

[D H]

My exact question is that what should I write instead of (d*2r/dt*2) in the formula?
Really what should I write?

I assume you meant $d^2r/dt^2$ here.

You will not get the desired answer if you consider r to be the scalar radial distance from the center of the planet. For example, r is constant for a circular orbit. It's derivative (and hence second derivative) is identically zero. You also will not get the desired answer for a non-circular orbit.

You need to consider r to be a vector, in which case $d^2\vec r/dt^2$ is acceleration.

 

[Cuttlas]

Thanks For your helps friends.

Now as the last question! While I was researching on this problem I found this:

F=-G.Me.m/r*2, you can see a negative sign behind the formula, but mine does not have it.

So what is this negative sign for? Is there any differences between (G.Me.m/r*2) and (-G.Me.m/r*2) ?

 

[D H]

It's to denote direction. However, making the scalar negative is a bit of an abuse of notation (in my opinion). A better way to denote direction is to write the force as a vector.

Denoting the displacement vector from the center of the Earth to the satellite as $\vec r$ and the unit vector in the direction of $\vec r$ as $\hat r$, the force vector per Newton's law of gravitation is
$$\vec F = -\frac{GM_em}{||\vec r||^2}\hat r$$
That unit vector can be removed by rewriting the above as
$$\vec F = -\frac{GM_em}{||\vec r||^3}\vec r$$

 

[andrien]

that negative sign in a simple sense means attractive force and so the orbit is a closed one(ellipse) otherwise it would be open like hyperbola when the negative sign is not there.