Homework Help: Kepler's law of periods problems about orbiting satellite

1. Nov 16, 2015

placebooooo

1. The problem statement, all variables and given/known data
Question 1: A orbiting satellite stays over a certain spot on the equator of (rotating) Pluto. What is the altitude of the orbit (called a "synchronous orbit")? (The radius of Pluto is 1150 km.)

Question 2:A orbiting satellite stays over a certain spot on the equator of (rotating) Venus. What is the altitude of the orbit (called a "synchronous orbit")? (The radius of Venus is 6050 km.)

2. Relevant equations
Kepler's law of periods:

T^3 = (r^3)(4pi^2)/GM

3. The attempt at a solution

Hi everyone, these two problems have been boggling me so bad. This is part of some online hw due midnight and these are the only two problems I was not able to do. I also have only one submission for each as I have no idea what I am doing wrong. I searched the forums and found a very similar question with a different planet and followed their work, but was too afraid to punch in my answer. I would love it if someone can confirm my work for me please. Here is my work:

T^3 = (r^3)(4pi^2)/GM

I solved for r:

r = [ (T^2)(G)(M)/(4pi^2) ]^1/3

For pluto, my first question, I used G = 6.67x10^-11, T = 6.39 days = 552096, and Mass of pluto, M= 1.309x10^22kg. I plugged in all of these values into my previous equation and got:

r = 18890566.49 m = 18890.56km

Then, h = r - r pluto = 18890.56 - 1150 = 17740.56km (final answer).

For venus, I did the same but instead I used the following values:
T = 116days and 18 hours = 10264800seconds
M =4.867x10^24 kg

Do the same as above, I got r = 953329010.5 m = 953329km

h = r - r venus = 953329km - 6050km = 947279km

Each question is qorth 13% of this hw assignment and would love it if someone could confirm whether these are correct or not. I really appreciate it.

2. Nov 16, 2015

Staff: Mentor

Your Pluto numbers look okay to me. For Venus however, your number of seconds for the period looks a bit high. 1.026 x 107 seconds is nearly 119 days.

3. Nov 16, 2015

placebooooo

Thanks for the reply. For Venus, I looked up that 1 day for venus is equivalent to 116 earth days and 18 hours. I converted this to seconds and used that for my value of T?

4. Nov 16, 2015

Staff: Mentor

Hmm. I'd be curious to know where you got that value. The sidereal rotation period of Venus is about 243 days long, or 5832.6 hours more precisely. (Venus' "day" is actually longer than its year! And thus the axial rotation is retrograde).

Either way, 116 days + 18 hrs would yield 10087200 seconds, so your calculation of that value is off. But I think you need to discard that value anyways since it doesn't reflect the rotation period of Venus.

5. Nov 16, 2015

placebooooo

Interesting! But very hard to picture indeed! I did not know that its year is shorter than its day. I learned a rather interesting fact today :)

Going back to the value, I got that value from google simply by searching "what is the mass of venus" and I got additional information on Venus on the sidebar of google :/

Anyway, using the alue for the sidereal rotation period for venus, 5832.6 hours yields 20997360seconds for T which ultimately gives me an altitude of 1530165.9km. Does that look any better?

6. Nov 16, 2015

Staff: Mentor

Ouch! If you search instead for "Venus rotation period" it returns 243 days. So. I guess that tells us something about trusting web resources.
Yup. Looks much better. Be sure to round to an appropriate number of significant figures.

7. Nov 16, 2015

placebooooo

That was definitely my fault. Definitely have to work on my google-Fu skills :)
Thank you so much for the help gneill! I greatly appreciate it!

8. Nov 16, 2015