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Universal gravitation 8- determine the speed of a satellite

  1. Jun 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the speed of a satellite moving in a stable orbit about the Earth if the satellite is 525 km above the Earth's surface.

    2. Relevant equations
    I have made a list of equations that are relevant for this entire module on universal gravitation. So although there are many of them does not mean that they all apply in this circumstance. The ones relevant to this question will be placed in bold.

    Kepler's 3rd law: (Ta/Tb)2=(Ra/Rb)3

    motion of planets must conform to circular motion equation: Fc=4∏2mR/T2

    From Kepler's 3rd law: R3/T2=K or T2=R3/K

    Gravitational force of attraction between the sun and its orbiting planets: F=(4∏2Ks)*m/R2=Gmsm/R2

    Gravitational force of attraction between the Earth and its orbiting satelittes: F=(4∏2Ke)m/R2=Gmem/R2

    Newton's Universal Law of Gravitation: F=Gm1m2/d2

    value of universal gravitation constant is: G=6.67x10-11N*m2/kg2

    weight of object on or near Earth: weight=Fg=mog, where g=9.8 N/kg
    Fg=Gmome/Re2

    g=Gme/(Re)2

    determine the mass of the Earth: me=g(Re)2/G

    speed of satellite as it orbits the Earth: v=√GMe/R, where R=Re+h


    period of the Earth-orbiting satellite: T=2∏√R3/GMe

    Field strength in units N/kg: g=F/m

    Determine mass of planet when given orbital period and mean orbital radius: Mp=4∏2Rp3/GTp2


    3. The attempt at a solution

    v=?
    Re=6.38x106m
    h=525km=525000m
    G=6.67x10-11N*m2/kg2
    me=5.98x1024kg

    R=Re+h=6905000m

    With all this information, I used to plug into the equation highlighted above to solve for v=7600.32m/s

    If someone could please have a look at my work and let me know if I made any mistakes, if so point them out to me! it would be greatly appreciated! Thank you so much in advance:)
     
  2. jcsd
  3. Jun 15, 2012 #2
    Correct again!
    IMHO I still suggest to start from basic equations like F=ma, F = GMm/R[itex]^{2}[/itex] and centripetal acc = mv[itex]^{2}[/itex]/R to derive the required equation for the speed v rather from attempting to remember the formula for v.
    Also I would have written v=√GMe/R, as v=√(GMe/R,).
     
  4. Jun 15, 2012 #3
    Thank you!
     
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