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Acceleration of a satellite

  1. Jul 6, 2008 #1
    A satellite is designed to orbit earth at an altitude above it's surface that will place it in a gravitational field with a strength of 4.5N/kg


    b)assuming the orbit is circular, calculate the acceleration of the satlellite and it's direction.

    since the satellite is rotating around the earth i used circular motion formula: a=v^2/r but since i wasn't provided any velocity i have to derive from the formula but i don't know where to start can i get some help please? thanks
     
  2. jcsd
  3. Jul 7, 2008 #2
    hey carbon,
    I am really exited to reply to this question. This is actually my first time on this site.

    to start off, the answer is right in front of you. A Force is defined as a Mass times an Acceleration. If F=m*a then a=F/m correct?
    voila!
    you have the acceleration. a=Newtons/kilograms. 4.5N/kg is really the acceleration because it is a= 4.5N/1Kg. do you see that?

    For the second part, to find the direction you do not need the formula a=v^2/r. Heck, you do not need this formula for the entire problem. All you need to realize is that the direction of the acceleration is ALWAYS towards the point where it is being accelerated towards. In this case, the point being accelerated towards is the center of earth. THE ACCELERATION in this problem is always DIRECTED towards the CENTER OF EARTH!!!!!.

    if you have anymore question on this or any difficulty understanding what i wrote, just let me know. It has been a pleasure
    -Alex
     
  4. Jul 7, 2008 #3
    i don't think the answer is 4.5N/Kg. they woudn't give me the answer in the question and then ask me to calculate it, wouldn't make sense. and the mass of the satellite is not given, so F=ma is not applicable in this case.
     
  5. Jul 7, 2008 #4
    I also thought the same at first but tried using different formulas but there were too many variables. I would also like to mention that a force is not expressed is N/kg. N/kg is an ACCELERATION. A force is expressed in N. To show you an example using F=m*a, N=kg*(N/kg) but not (N/kg)=kg*(N/kg) . So the 4.5 that you gave HAS to be the acceleration because it is expressed in N/kg.

    to use the equation F=G*m1*m2/(r^2) to find the acceleration, you would need to know r.

    to use a=v^2/r you would need to know either the time of orbit or the velocity and the radius

    to use F=m*a you would need to know the mass of the satellite

    Do you see my point, if you said that 4.5N was the force, none of these would work with the given info. You need more information to solve this problem.

    Make sure that the source where you got the problem from mentions the mass of the satellite in previous pages.
     
  6. Jul 7, 2008 #5

    kreil

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    [tex] g = \frac{GM}{r^2}[/tex]

    Using the above equation, with g=4.5 N/kg, G=gravitational constant, M= earths mass you can calculate the altitude the satellite would have in order for it to experience that specific gravitational force. This is R.

    Now, in order for the satellite to orbit the planet and not just fall down, you need to balance the forces:

    [tex]F_{net}= \frac{M_{sat} v^2}{R}[/tex]

    [tex]F_{grav}= \frac{M_{sat} M_{earth} G}{R^2}[/tex]


    [tex]\frac{M_{sat} v^2}{R}=\frac{M_{sat} M_{earth} G}{R^2}[/tex]

    [tex]v^2=\frac{M_{earth} G}{R}[/tex]

    As you can see, the mass of the satellite cancels out so it is irrelevant.

    Now you have the velocity and the radius so you can calculate the acceleration using your a=vv/r equation.
     
    Last edited: Jul 7, 2008
  7. Jul 7, 2008 #6

    D H

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    Newtons/kg are not primitive units (i.e., units expressed as the produce of powers of mass, length, time, electrical charge). What is 4.5 N/kg in primitive units?
    The satellite's mass is not needed to solve this problem.
     
  8. Jul 7, 2008 #7

    Doc Al

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    It really is as simple as that. (Just as tiale11 described in his first post.) Of course you'd express the acceleration in its normal units.
    Just call the mass of the satellite M. You don't need the actual value.

    I see that this is part "b" of the problem. What was part a?
     
    Last edited: Jul 7, 2008
  9. Jul 7, 2008 #8

    kreil

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    I don't think it is that simple...if that were the acceleration of the satellite then that would imply it has no orbital velocity, and hence falls straight back down to earth.
     
  10. Jul 7, 2008 #9

    Doc Al

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    Why in the world would you think that? The acceleration still equals [itex]v^2/r[/itex].

    But since the gravitational force is proportional to the mass, this is the acceleration. Note that near the earth's surface, the gravitational field strength is called g, which equals 9.8 N/kg = 9.8 m/s^2.
     
  11. Jul 7, 2008 #10

    kreil

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    Notice that 4.5N/kg is the strength of the gravitational field, not the acceleration of the satellite. My post above(#5) outlines how to balance the forces to find the acceleration of the satellite.
     
  12. Jul 7, 2008 #11

    kreil

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    So what you're saying is that my post (#5) is incorrect? I find this hard to believe...
     
  13. Jul 7, 2008 #12

    D H

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    4.5 N/kg is the acceleration of the satellite. The only force acting on the satellite is gravity; there are no forces to balance.
     
  14. Jul 7, 2008 #13

    Doc Al

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    It's all unnecessary. F=ma works just fine. F = M(4.5 N/kg) = Ma, thus a = 4.5 N/kg (or m/s^2). (Gravity is the only force acting on the satellite.)
     
  15. Jul 7, 2008 #14

    kreil

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  16. Jul 7, 2008 #15

    Doc Al

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    There's only one force acting, but OK.

    Fine. Now divide by M_sat:

    [tex]a = \frac{v^2}{R}=\frac{M_{earth} G}{R^2}[/tex]

    Realize that the right hand term is the gravitational field strength, as given.

    You don't need to solve for the speed.
     
  17. Jul 7, 2008 #16

    D H

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    Hint: The newton is not a primitive unit. Rather one newton is defined as the force needed to make a mass of one kilogram accelerate at one meter per second squared. Applying Newton's second law, [itex]1\,\text{newton} = 1\,\text{kg}\cdot\text{meter}/\text{second}^2[/itex]. What is [itex]1\,N/kg[/itex]?
     
  18. Jul 7, 2008 #17

    Doc Al

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  19. Jul 7, 2008 #18

    kreil

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    This makes it more clear.

    touche
     
  20. Jul 7, 2008 #19
    hey kreil,
    you really did not answer the original question. its nice that you derived the formulas but how exactly do they help you? you still have 3 unknowns: the velocity, acceleration, and the radius. a N/kg is not a centripetal force but is the centripetal acceleration. The answer to the original question is a=4.5m/s^2 and the direction is towards the center of the Earth. No computations are needed all that is needed is to know how to read UNITS. Thats it!!!!!
     
  21. Jul 7, 2008 #20
    no tiale11 a calculation is needed. the question ask to "calculate the acceleration of the satellite and it's direction" the question i'm asking is how to munipulate the centripetal formula
     
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