# Using Taylor expansion and Kepler's Law to derive approximation for orbital period

1. Sep 13, 2009

### ArbazAlam

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

Use Kepler's Third Law and a Taylor expansion to derive the following approximation for the orbital period of a satellite in low Earth orbit with a constant height h above the surface of the Earth. h << R_earth :

$$P \approx P_{0}(1+3h/2R_{e})$$

2. Relevant equations

Kepler's Third Law:

$$P^{2}=4\pi^{2}r^{3}/GM$$

Taylor Expansion:

$$f(x)=\sum^{\infty}_{n=0}f^{n}(x-a)/n!$$

3. The attempt at a solution

I'm pretty certain my professor doesn't want us to expand the Taylor series beyond the first degree. I tried to write Kepler's Third Law in terms of just P and then write out the Taylor expansion about 0 (McLaurin series), but all terms came out to 0. I'm not 100% sure where to go beyond the definitions of Kepler's Law and the Taylor expansion.

I'm also confused why, if we are assuming h<<R_e, the variable h is even necessary.

I'm not asking for a solution, but any hints are much appreciated.

Last edited: Sep 14, 2009
2. Sep 13, 2009

### D H

Staff Emeritus
Re: Using Taylor expansion and Kepler's Law to derive approximation for orbital perio

How are you going to get a square root in a Taylor expansion? Think about it.

(Note: That is a hint, not a question re the validity of the problem.)

3. Sep 13, 2009

### ArbazAlam

Re: Using Taylor expansion and Kepler's Law to derive approximation for orbital perio

Doesn't the original function have to have a root to begin with?

4. Sep 13, 2009

### D H

Staff Emeritus
Re: Using Taylor expansion and Kepler's Law to derive approximation for orbital perio

So? The Taylor expansion of

$$f(x) = \sqrt(1+x)$$

$$f(x) = 1+\frac 1 2 x - \frac 1 8 x^2 + \frac 1 {16} x^3 - \frac 5 {128} x^4 + \cdots$$

Note: There is no square root in the expansion. Truncating a Taylor expansion yields a polynomial. How do you go from a polynomial to a square root?

I'll let you think about it for a bit but will give you a more direct hint in the next post if you can't see how to go forward.

5. Sep 14, 2009

### ArbazAlam

Re: Using Taylor expansion and Kepler's Law to derive approximation for orbital perio

I made a dumb mistake. There is no root in the period approximation and I edited accordingly.

D H, I'm afraid I still don't understand. :( Thank you for bearing with me.

6. Sep 14, 2009

### D H

Staff Emeritus
Re: Using Taylor expansion and Kepler's Law to derive approximation for orbital perio

In the previous version of the problem you would have computed a truncated Taylor expansion of the square of the period and then taken the square root at the end.

This new form is even easier. Just do a Taylor expansion of the period, truncate to the linear term, and you are done. The problem pretty much tells you what to do a Taylor expansion about.

Do you understand the basic concept of how to compute a Taylor expansion?

7. Sep 14, 2009

### ArbazAlam

Re: Using Taylor expansion and Kepler's Law to derive approximation for orbital perio

Yes, I can compute a Taylor expansion like the example you presented.

So in this problem, I take f(h) = P^2 and then compute the Taylor series for f(h) about R_e [to n=1]. Once I do that, I can take the square root of the polynomial. Is this correct?

8. Sep 14, 2009

### D H

Staff Emeritus
Re: Using Taylor expansion and Kepler's Law to derive approximation for orbital perio

No, you take f(h) = P. Taking f(h)=P2 will lead you to a different approximation, one that involves a square root. There is no square root in the edited version of the problem.

9. Sep 14, 2009

### ArbazAlam

Re: Using Taylor expansion and Kepler's Law to derive approximation for orbital perio

Sweet! Finally got it. Here is how I got it for future reference.

I took the Taylor series of P(h+R_e) about R_e and got:

$$P\approx 2\pi R_{e}\sqrt{(h+R_{e})/(GM)}(1+(3h)/(2R_{e}))$$

The next question asks what the value of P_0 is. I take that to be:

$$P_{0}=2\pi R_{e}\sqrt{(h+R_{e})/(GM)}$$

The units work out and everything. Thanks a million, D H! If this forum had a reputation system, I'd give you tons.

10. Sep 14, 2009

### D H

Staff Emeritus
Re: Using Taylor expansion and Kepler's Law to derive approximation for orbital perio

Almost.

The Taylor expansion of some point f(x) about some point x0 is

$$f(x) = f(x_0) + (x-x_0)\frac {df}{dx}\Bigr|_{x=x_0} + \cdots$$

So what is this reference point in this problem? You are given height above the surface h, so the reference point is the Earth's surface. P0 is the period for an object that would orbit the right fractions of inches above the surface were it not for pesky big things like Mt. Everest and pesky little things like the atmosphere.

And thanks for the kudos.

11. Sep 16, 2009

### ArbazAlam

Re: Using Taylor expansion and Kepler's Law to derive approximation for orbital perio

Ok, so:

$$P(R_{e})=2 \pi \sqrt{R_{e}^{3}/(GM)} = P_{0}$$

$$P'(R_{e})=3 \pi \sqrt{R_{e}/(GM)}$$

So,

$$P(h+R_{e}) \approx P(R_{e}) + P'(R_{e})(h) = 2 \pi \sqrt{R_{e}^{3}/(GM)}(1+(3h)/(2R_{e})) = P_{0}[1+(3h)/(2R_{e})]$$

Last edited: Sep 16, 2009