Deriving Orbital Period with Kepler's & Taylor Expansions

[ArbazAlam]

Homework Statement

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 :

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

Homework Equations

Kepler's Third Law:

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

Taylor Expansion:

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

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.

 

[D H]

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.)

 

[ArbazAlam]

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

 

[D H]

So? The Taylor expansion of

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

about x=0 is

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

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.

 

[ArbazAlam]

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.

 

[D H]

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?

 

[ArbazAlam]

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?

 

[D H]

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.

 

[ArbazAlam]

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.

 

[D H]

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.

 

[ArbazAlam]

Ok, so:

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

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

So,

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