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    I need help on this question, I will attach a link to the screenshot:
    http://prntscr.com/akpown

    I thought of using Newton's law of gravitation, given by F = GMm/r^2
    However in the question they've provided us with the radius which the satellite orbits relative to the surface of the Earth; therefore should I add the radius of the Earth with the radius of the orbit?
    So what I mean is:

    F = (6.67*10^-11)(5.98*10^24 KG)(2.5*10^3 KG) /
    (1.6*10^7 + 6.4*10^6)^2

    Giving me a value of 1987N?
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    Hi,
    Yeah I think the radius of the Earth needs to be added- when I worked it out I got roughly 1990 N
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    (Original post by lharrisonx)
    Hi,
    Yeah I think the radius of the Earth needs to be added- when I worked it out I got roughly 1990 N
    Thanks for the response, however I went ahead and looked at the answer and they did not consider the radius of the Earth in the calculation and had gotten a value of 3910N, could they potentially be incorrect?

    This is where I found it:
    http://www.physbot.co.uk/uploads/1/2...ity_aqa_ms.pdf

    If you scroll down to page 13/16 on the PDF viewer.
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    No you do not add the radius of earth.

    Think about "an orbit of radius xxx". What does that actually mean?

    It means it moves in a circular path (key word there being radius). So if it's moving in a circular path around the earth, then they must have already taken in to account the radius of earth.



    Also this is incorrect
    (Original post by 1017bsquad)
    However in the question they've provided us with the radius which the satellite orbits relative to the surface of the Earth
    It's not relative to the surface because it's already a radius
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    (Original post by Student403)
    No you do not add the radius of earth.

    Think about "an orbit of radius xxx". What does that actually mean?

    It means it moves in a circular path (key word there being radius). So if it's moving in a circular path around the earth, then they must have already taken in to account the radius of earth.



    Also this is incorrect


    It's not relative to the surface because it's already a radius
    Ahh I see now, thanks for clarifying, I should really pay attention next time, thanks though I appreciate it!
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    (Original post by 1017bsquad)
    Ahh I see now, thanks for clarifying, I should really pay attention next time, thanks though I appreciate it!
    Most welcome
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    Oops, I really need to read questions more carefully Sorry!
 
 
 
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