Solving Kepler's Third Law for Period - 6.7E6m

[KVat390]

Homework Statement

Find the period of a satellite that is in orbit 6.7×10^6 meters from the center of the earth?

Homework Equations

P^2=R^3 P=period and R=average distance

The Attempt at a Solution

so far I have tried P^2=(6.7×10^6)^3
then sqrt of (P^2)=sqrt of (3.00763×10^20)
My answer is the period of the satellite=1.73425×10^10
I'm pretty sure my answer is way off, please describe how to get the correct answer.

 

[jedishrfu]

I think you need to use the formula 4*pi / T^2 = G*M / R^3 where M is the Earth's mass, T is the period and R is the distance between the two bodies.

http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion

see the section on the third law

 

[KVat390]

4∏/T^2=GM/R^3
4∏/T^2=(6.67×10^-11)(5.98×10^24)/(6.7×10^6)^3
12.57/T^2=(3.98866E14)/(3.00763E20)
12.57/T^2=1.326180414E-6
T^2=1.66700878E-5
T=.0040829019
Is this answer correct?

 

[jedishrfu]

I don't think so. How did you get T^2= ?

I would have solved for T^2 first and then inserted the values. It always makes things easier.

GM*T^2 = 4*pi^2 * R^3

T^2 = 4*pi^2*R^3 / G*M

 

[SteamKing]

Unit-free calculations almost never end well.

 

[KVat390]

A Revision to my previous post

The equation is 4pi^2/T^2=GM/R^3 not 4pi/T^2=GM/R^3
I changed my answer to 5.46 periods
39.48/T^2=(3.98866E14)/(3.00763E20)
T^2=39.48/1.33x10^-6
sqrt of T=sqrt of 39.48/1.33x10^-6
T=5.45×10^3 periods/earth years
Is my answer correct?

 

[jedishrfu]

The equation is 4pi^2/T^2=GM/R^3 not 4pi/T^2=GM/R^3
I changed my answer to .00723578554
39.48/T^2=(3.98866E14)/(3.00763E20)
T^2=5.235760276E-5
sqrt of T=sqrt of 5.235760276E-5
T=.0072358554 periods/earth years
Is this answer correct?

I fixed my eqns to say pi^2

so that T^2 = 4 * pi^2 * R^3 / (G * M) = 4 * pi^2 (6.7E6)^3 / ((6.67E-11) * (5.98E24)) = 2.98E7

and T =5.46E3

When I did my calculations I got tripped up by a lack of parentheses around the denominator terms which meant that I divided by the G and then multiplied by the M instead of dividing by (G * M)

 

[SteamKing]

Still won't knuckle under and use units, eh? Good Luck with the rest of your course work!

 

[jedishrfu]

Still won't knuckle under and use units, eh? Good Luck with the rest of your course work!

Klingon programmers have no need of units, we use junits...

 

[jedishrfu]

Edit: To the OP I forgot the pesky pi factor again but fixed my post above.

Having gotten the answer for T and knowing R does it seem reasonable?

What object might have a periodicity like this?

Oh and as SteamKing said where is your unit analysis? What units is used for T?

 

[D H]

I fixed my eqns to say pi^2

so that T^2 = 4 * pi^2 * R^3 / (G * M) = 4 * pi (6.7E6)^3 / ((6.67E-11) * (5.98E24)) = 9.47E6

and T = 3.07E3

That result is incorrect. You dropped a factor of pi.

You are being sloppy and it is getting you in trouble.You also are not following instructions. You didn't read the instructions on where to post your physics problems.

If your thread mysteriously disappears without a trace from this forum, you probably didn't post it in the right place, so your thread was moved to the Introductory Physics or Engineering forum.

Advanced physics does not mean, among other things:

• "I think this problem is really hard."
• "I'm taking a college physics course."
• "We're covering electromagnetism now."
• "We're covering relativity now."
• "We're covering quantum mechanics now."

If you have to plug numbers into a formula to get the final answer, that's usually a good sign the question doesn't belong here.

Questions that do belong in this forum are from upper-division or graduate physics courses — classes physics majors (in the US) take in their junior year and later.

This is not an advanced physics homework problem. I'm moving this to the introductory physics homework section.

 

[SteamKing]

'Advanced Physics' also doesn't mean, "Units? Units? I don't need no stinkin' units!"