1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Circular Orbits of Satelites and Planets

  1. Jun 22, 2008 #1
    1. The problem statement, all variables and given/known data

    A spy satellite is in circular orbit around Earth. It makes one revolution in 6.00 h.

    A. How high above Earth's surface is the satellite?

    B. What is the satellite's acceleration?


    2. Relevant equations

    I'm not sure but I believe we could possibly derive something from

    SUMofF = (Gm1m2)/r^2

    and with NSL, v=sqrtof (GMe/r)

    where Me is the mass of the Earth 5.974e24 kg. G is the universal gravitational constant 6.67e-11N * m^2/kg^2




    3. The attempt at a solution

    and r is the radius of the earth plus the radius of the height. So we can find the height that the satellite is by deducting the radius of the earth (6.371e6 m) from r to find the height.

    I don't know what equation to use and I don't know how I would find the acceleration.


    Thanks for your help!
     
  2. jcsd
  3. Jun 22, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

    The force of attraction between the Earth and satellite,provides the force needed to keep the satellite in a circular orbit.
     
  4. Jun 22, 2008 #3
    would it be the Universal Gravitational Constant? Sorry, I'm not really clear with what you're saying. Thanks
     
  5. Jun 22, 2008 #4

    dynamicsolo

    User Avatar
    Homework Helper

    If you set the gravitational force between Earth and satellite, given by

    equal to the centripetal force acting on the satellite in its circular orbit (i.e., gravity is what provides the centripetal force), and do some algebra, you'll come up with

    This is called the "circular velocity" of the spacecraft in its orbit. If you now consider that the period, T, of the satellite is the time it takes for the spacecraft to complete the circular orbit of radius R at the speed v given above, you can do some more algebra to find a relation between T and R (you will, in fact, have found a version of Kepler's Third Law!).

    The gravitational force equation will give you the satellite's acceleration in its orbit, if you keep in mind Newton's Second Law (F = ...?). [Since what they are asking for is also the centripetal acceleration of the satellite, you can use that equation and the orbital speed you found. You should -- you'd better! -- get the same answer either way...]
     
  6. Jun 22, 2008 #5
    Hi again dynamic solo, thanks for all the help lately. Before I go further, r in the "circular velocity" equation is the radius of earth? thanks
     
  7. Jun 22, 2008 #6

    rock.freak667

    User Avatar
    Homework Helper

    r is the distance between the earth and the satellite.
     
  8. Jun 22, 2008 #7
    Thank you. So for V, that would be 2pi/6h?
     
  9. Jun 22, 2008 #8

    dynamicsolo

    User Avatar
    Homework Helper

    What is the distance around the circular orbit? That is the distance you are covering in those six hours.
     
  10. Jun 22, 2008 #9

    The distance is 2pi, isn't that right? so 2pi/6h?
     
  11. Jun 22, 2008 #10

    dynamicsolo

    User Avatar
    Homework Helper

    The circumference of a circle is given by what formula?

    (The number pi has no units, so 2·pi can't be the length of anything...)
     
  12. Jun 22, 2008 #11
    hi dynamic,

    the circumference is 2pi*radius

    but we don't know the radius. So I have


    2pi*radius/6hours = sqrtof(GMe/r)




    solving for r I get:


    r ^1.5 = sqrt of (GMe) * 6 all over 2pi


    oh god, what am I doing?
     
  13. Jun 22, 2008 #12

    dynamicsolo

    User Avatar
    Homework Helper

    OK, the circular speed is given by the distance around the circular orbit (circumference) divided by the time required to do so (period), so you get

    But watch your algebra! You have r on one side and a square root of r in the denominator of the other side. Square both sides (also convert the period into seconds, since we'll need to work in SI metric) to get

    [ (2·pi) · r ]^2 / (6 hours converted to seconds)^2
    = G · Me / r .

    Now solve this for r , which is the radius of the orbit. (Actually, what you had is close, but r^1.5 will be more of a nuisance to work with...)
     
    Last edited: Jun 22, 2008
  14. Jun 22, 2008 #13
    hey dynamic, once again, thanks a lot for the help.

    I got it right I think!

    1.039E7 meters!!!

    hehehe
     
  15. Jun 22, 2008 #14

    dynamicsolo

    User Avatar
    Homework Helper

    ...or about 10,400 km. altitude.

    Now how about part (b)?
     
  16. Jun 23, 2008 #15
    hmmm i needed help with that,

    I got 8.7E-1 meters/second


    that's wrong huh?

    thanks
     
  17. Jun 23, 2008 #16

    dynamicsolo

    User Avatar
    Homework Helper

    Well, acceleration has units of meters/(second^2), so something didn't come out right...

    You should probably show how you calculated this. In the meantime, here are two approaches you can try (there's a third, but I'll describe it afterwards):

    You know that the force on the satellite is F = ma , so you could compute either

    1) the gravitational force between the satellite and the Earth, set that equal to ma , and solve for a

    or

    2) find the centripetal force on the satellite (you have the radius of the orbit and the speed of the satellite along it) and set that equal to ma, and solve for a .

    Notice that you don't need to know the satellite's mass m, since it will divide out (which tells us that any satellite which is small in mass compared to Earth would behave the same way).
     
  18. Jun 23, 2008 #17
    Hi Dynamicsolo,

    yes it is 8.7E-1 m/s^2

    this is what I did.

    I used the formula asubr = omegasquared*radius

    i found the requency and plugged that into the equation

    arrived with that answer. The problem is, I don't know if I did it 100% correctly.

    I'll try your method and report back. Thank you greatly
     
  19. Jun 23, 2008 #18
    I'm a bit confused with steps 1 and 2 since there are many names for the same thing. First off, what do you mean "gravitational force" is that the same thing as Fg = Gm1m2/r^2??


    and with what you're saying, we don't need to know the mass of the satellite therefore:

    Fg= GMe/r^2 ???

    and then plugging in the universal constant and multiplying it with the mass of earth and dividing everything by the radius of the orbit squared will give us the gravitational force? Thanks


    edit -

    and since you said


    gravitational force = m * a

    what is the value of m? which mass is it? thanks
     
  20. Jun 23, 2008 #19

    dynamicsolo

    User Avatar
    Homework Helper

    Ah, that form for the centripetal acceleration will work too. It also looks like you found the angular frequency of the satellite on its orbit correctly. But, did you use the radius of the orbit, or the altitude...?
     
  21. Jun 23, 2008 #20

    dynamicsolo

    User Avatar
    Homework Helper

    That is the equation for the gravitational force between two masses.

    You left out one of the masses this time. The gravitational force between the Earth (mass: Me) and the satellite (call its mass m) will be

    F_g = G · Me · m / (r^2) ,

    which will equal the centripetal force on the satellite (since the gravitational force is what is providing that force)

    F_g = F_c = m · a_c ,

    the centripetal acceleration being what you are solving for. The mass m appears on both sides of the equation, so it divides out (meaning that, for satellites with masses that are small compared to Earth, the mass doesn't matter -- we must modify this approach to deal with, say, the Moon).
     
  22. Jun 23, 2008 #21

    BobG

    User Avatar
    Science Advisor
    Homework Helper

    You wanted acceleration, not force, so you divided the mass out of the equation. So, instead of
    Fg= GMe/r^2 ???

    it should be:

    ag = GMe/r^2

    which is what you were trying to find.
     
  23. Jun 23, 2008 #22
    ok thanks everyone for your help. I appreciate it.
     
  24. Jun 23, 2008 #23

    dynamicsolo

    User Avatar
    Homework Helper

    I want to mention something further about finding the satellite's acceleration. You used the form of the centripetal acceleration equation involving the angular frequency of the orbit to get

    a_c = (2·pi/21600 sec)^2 · 1.676·10^7 m = 1.42 m/(sec^2) .

    You could also have used the version involving the linear velocity on the orbit:

    v = C/T = (2·pi·r)/T or sqrt[G·(Me)/r] = 4875 m/sec , and

    a_c = (v^2)/r ,

    which is otherwise equivalent to the form you used.

    You could also use the gravitational force equation, to find

    F_g = G·Me·m/(r^2) = F_c = m·a_c ,

    m being the satellite's mass [this equation, BTW, also gives us the "weight" of the spacecraft at this altitude above the Earth's surface -- more on this presently] , which leads to

    a_c = G·Me/(r^2)
    = [6.67·10^-11 N·(m^2)/(kg^2)] · (5.98·10^24 kg) / [(1.676·10^7 m)^2]
    = 1.42 m/(sec^2) .

    Since m does not appear in this result, this shows that any small satellite experiences the same acceleration on this orbit.

    There is yet another approach using comparison ratios. The gravitational force between Earth and the satellite obeys the same relation at any radius from the Earth's surface outward. So the acceleration the satellite would experience at the Earth's surface would be

    a' = G · Me / (R^2) ,

    where R is the Earth's radius. You may recognize this expression as giving g , the acceleration of gravity at the Earth's surface. We can now compare this to the acceleration in the satellite's orbit:

    a_c / g = [ G·Me/(r^2) ] / [ G·Me/(R^2) ] = (R/r)^2 .

    In this method, we don't even need to know the value of G or the Earth's mass (much less the satellite's mass, m), since the constants all cancel out in the comparison. So we find

    a_c = [ (R/r)^2 ] · g = [ (6378 km / 16,758 km)^2 ] · (9.81 m/sec^2)

    = (0.3806^2) · (9.81 m/sec^2) = 1.42 m/(sec^2) .

    This is another way of saying that the satellite in orbit weighs 0.3806^2 = 0.145 (about 1/7) of what it weighs on Earth.

    Many roads leading to the same place (because they're all really basically equivalent)...
     
  25. Jul 13, 2008 #24
    Wow, thanks a lot for explaining that clearly dynamicsolo, I appreciate it!!! That's helpful to know. :)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook