Mass of a planet given it's satellites orbital radius & period

  • Thread starter bbhh
  • Start date
  • #1
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Ah, please, someone help me. I've been working on this problem for half an hour, it's the last question on my review before my test, which i want to write right now (correspondence).

Homework Statement



If a satellite requires 2.5 h to orbit a planet with an orbital radius of 2.6 x 10^5 m, what is the mass of the planet?


Homework Equations



ac = v^2/r = 4 pi^2 r / T^2

v = sqrt(GM / r)

(... at least that's what i think?)

The Attempt at a Solution



1. I attempted to find the velocity from the radius (2.6*10^5) and the time (2.5hr*60*60=9000s)

v^2/r = 4 pi^2 r / T^2
v^2 = 4 pi^2 r^2 / T^2

v^2 = 4 * 3.14^2 * 2.6*10^5 / 9000^2
v=.356m/s

2. I used v= sqrt(GM/r) to find the mass.

.356= sqrt(6.67*10^-11m/2.6*10^5)
.127=6.67*10^-11m/2.6*10^5
330200=6.67*10^-11m
m= 5*10^15kg

the answer key says it's 1.3*10^20kg
 
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Answers and Replies

  • #2
The answer key is correct, I am unsure about what mistake was made in the attempted solution...maybe a calculator entering issue...
v^2/r==GmP/r^2
4pi^2/T^2==GmP/r^3
4pi^2r^3/T^2*G==Mp
Mp==1.284x10^20 kg.
 
Last edited:
  • #3
23
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i appreciate your help but i think i need a bit more in depth of an explanation.

"v is actually the square root of GMp/r"... isn't that what i did?... see... a little lost.
 
  • #4
i appreciate your help but i think i need a bit more in depth of an explanation.

"v is actually the square root of GMp/r"... isn't that what i did?... see... a little lost.
Yes, when I reread your post, that is what you did...I don't know why you didn't get the right answer...However, if you follow the steps I showed in my previous post, you will get the correct answer.
 
  • #5
23
1
hmm thanks. i wonder if i just screwed up my maths somewhere along the way. i checked it so many times though... weird
 

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