Orbit height around planet

  • Thread starter Pseudopro
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  • #1
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Homework Statement


I have been able to complete section a and b but c doesn't match the answer

a)The acceleration of free fall at the surface of a planet is g and the radius of the planet is R. Deduce that the period of a satellite in a very low orbit is given by T=2pi sqrt(R/g).

b)Given that g=4.5ms^-2 and R=3.4x10^6m, deduce that the orbital period of the low orbit is about 91 minutes.

c)A spacecraft in orbit around this planet has a period of 140 minutes. Deduce the height of the spacecraft from the surface of the planet. (Ans: 3.1x10^6)

Homework Equations


T=2pi sqrt(R/g)
T=2pi sqrt(R^3/GM)

The Attempt at a Solution


I can't obtain the answer 3.1x10^6 in c). When doing c), T=2pi sqrt(R/g) can't be used with g=4.5ms^-2 because period of 140 minutes means acceleration of free fall has already changed so I used T=2pi sqrt(R^3/GM):
140x60=2pi sqrt((R+h)^3/GM) {used R+h because the question wants height of spacecraft from SURFACE of planet}. Now I must find M. Using information obtained from b): 91x60=2pi sqrt((3.4x10^6)^3/GM), M=7.81x10^23. Substituting M into 140x60=2pi sqrt((R+h)^3/GM), I get R+h=4.53x10^6, h=1.13x10^6!!??
I've thought this over for a very long time. Have I misunderstood the question or is my method wrong? Someone help me! Thank you!
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi Pseudopro! Welcome to PF! :smile:

(have a pi: π and try using the X2 iocn just above the Reply box :wink:)
… When doing c), T=2pi sqrt(R/g) can't be used with g=4.5ms^-2 because period of 140 minutes means acceleration of free fall has already changed so I used T=2pi sqrt(R^3/GM):
Why are you using G ? :confused:

You know g at distance R, and you know how gravity depends on distance, soooo … ? :wink:
 
  • #3
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Hi Pseudopro! Welcome to PF! :smile:

(have a pi: π and try using the X2 iocn just above the Reply box :wink:)


Why are you using G ? :confused:

You know g at distance R, and you know how gravity depends on distance, soooo … ? :wink:
Yes, you're right - g of 4.5 is at distance R, and gravity changes with distance. My way of calculating the new g at height h was by using g=GM/R2. This is why I used G - it shouldn't be like that? This is what I did:
4.5=6.667x10-11M/(3.4x106)2
M=7.81x1023
gnew=6.667x10-11M(just calculated)/(R+h)2
g(R+h)2=5.20x1013...(1)

T=2π sqrt((R+h)/g) [cleaner way to write square root?]
140x60=2π sqrt((3.4x106+h)/g)
Cleaning up: (3.4x106+h)/g=1.79x106...(2)

(1)x(2): (3.4x106+h)3=9.31x1019
Solving: h=1.13x106

If the book's answer 3.1x10^6 is correct, there is something wrong with my method. Please tell me if this (use of g=GM/R2) is incorrect. Also could you verify that my use of R+h is correct? Thank you very much for helping me :smile:
 
  • #4
tiny-tim
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Homework Helper
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Hi Pseudopro! :smile:

Your method is very long-winded. :redface:

You started with T=2π √(R3/GM) …

in other words, R is proportional to T2/3 ……

that is all you need to know!!
If the book's answer 3.1x10^6 is correct, there is something wrong with my method. Please tell me if this (use of g=GM/R2) is incorrect. Also could you verify that my use of R+h is correct? Thank you very much for helping me :smile:
I get the same R+h as you do.

I think you'll find the question has a misprint, and it should be 240 minutes, not 140. :wink:
 
  • #5
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in other words, R is proportional to T2/3 ……

that is all you need to know!!
Oh! that's how you do it! I do have a knack of making simple questions complicated but still reaching the solution. Thank you very much!
 
  • #6
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It's also possible they did R to T3/2, just realised
 
  • #7
gneill
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You can dispense with G and M altogether if you wait a step or two before plugging in values.

Suppose GM = µ. Then g = µ/R2, or rearranged, µ = gR2.

You're given one set of values for g and R, so for another radius r,

gr = µ/r2 = gR2/r2

That's the inverse square law at work!
 

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