Question

Earth has a mass of 5.97 x 10^24kg, and a mean radius of 6.38 x 10^6m. What would be the orbital speed and period of a satellite in orbit 1.44 x 10^8 m above Earth?

Answer 1

@d3Xter There is multiple choices but the copy and paste wont work on their certain text.

Answer 2

circular orbital motion is independent of the mass of the orbiting body

Answer 3

well, is there anyway to screenshot my screen and put it up for you to see? because the answers look like you needa do math.

Answer 4

Just a minute

Answer 5

found a way

Answer 6

I am on a time limit as well :) no hurry or anything ;)

Answer 7

\[P=\sqrt{\frac{ r^34 \pi ^2 }{ Gm _{1} }}\]

Answer 8

I appreciate your enthusiasm in helping me understand how to do it. but this is my last credit to graduate highschool and i am in no hurry to understand it.

Answer 9

ah well - maybe we're in no hurry to help
Most helpers here have a true interest in the subject and in the learning process.

Why would anyone want to give an anser for you to copy out on a paper and forget ...?

Answer 10

Well obviously i do not care about what you have to say Mr. Nood. I do not believe i asked to learn it now did I? You don't have to help me. Your input to me is pointless and somewhat a pathetic waste of time. Excuse my ignorance but Sianora.

Answer 11

But thank you for wasting my time. You deserve a medal.

Answer 12

Thank you for your help @d3Xter I understand if you do not want to give me the answers anymore.

Answer 13

\[r = (R + h) \]where r = Radius of earth + Altitude of the orbiting planet

So \[r = 1.5038 \times 10^8 m\]Hence
\[P = \sqrt{\frac{ (1.5038 \times 10^8 )^3 \times 4 \pi^2 }{ 5.97 \times 10^{27} \times 6.67 \times10^{-11}}} \]\[P = 5.80 \times 10^5s\]
And then \[V = \frac{ 2pir }{ P }\]which gives us
\[V = \left( \frac{ 2 \pi \times 1.5038 * 10^8}{ 5.8 \times 10^5 } \right) m/s\]\[V=1629 m/s\]