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Need Check On Orbital Satellite Problem

  1. Dec 7, 2007 #1
    1. The problem statement, all variables and given/known data
    A 347 kg satellite orbits mercury and makes 6.6 orbits in 1 day.

    Mercury ---- Radius 2.44*10^6 meters ----- Mass 3.27*10^23 kg

    1.Find the Period of Revolution
    2.Calculate the Tangential Velocity
    3.Calculate the height above the surface of the planet that the satellite is in orbit
    4.Calculate the force of gravity at a distance of 1.50*10^6 meters

    2. Relevant equations
    Planetary motion Equations

    3. The attempt at a solution

    1. Period= 24 hours/6.6 orbits ---- 13090.90909 seconds (per revolution)

    2. Tangential Velocity= (2*pi*radius)/Period ----- (2*pi*(2.44*10^6))/13090.90 second ---- Vt= 1171 m/s

    3. My friend said that this was the equation for the third part, does anyone know if this is right.
    h=((Gravity Constant*Mass Mercury)/Vt^2) - radius of Mercury -----
    h=((6.67*10^-11)(3.27*10^23)/1171^2) - 2.44*10^6 meters -----
    h= 13465956 meters or 1.3465*10^7 meters (is this right)!?!

    4. Same friend said that the force of gravity would be :
    g= (Gravity Constant*Mass Mercury)/((Radius Mercury+Height)^2)
    g= (6.67*10^-11)(3.27*10^23)/ ((2.44*10^6+1.5*10^6)^2)
    g= 1.405 m/sec^2

    Can some one check these answers because I just feel like I did something wrong.
     
  2. jcsd
  3. Dec 7, 2007 #2
    Hello,

    You've used the radius to represent two different quantities.

    When you calculated the velocity in part (2), you used the radius 2.44*10^6 m as the radius of the orbit. However, when you calculated the height in part (3), you used the radius of 2.44*10^6 m as the radius of mercury. I'm not sure which it's supposed to be, but it can't be both.
     
  4. Dec 7, 2007 #3
    Can anyone else help?
     
  5. Dec 7, 2007 #4

    Dick

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    Listen to dotman first. Your result for the tangential velocity is correct only if the radius of Mercury is the same as the radius of the orbit. You need to use 'Gravity constant' from the beginning. GM/r^2=v^2/r. v=2*pi*r/T. T is period, M is mass of Mercury and r is the radius of the orbit. The radius of Mercury doesn't matter much.
     
  6. Dec 8, 2007 #5
    I redid the questions

    1. Period= 24 hours/6.6 orbits ---- 13090.90909 seconds (per revolution)

    2. Book says : ((G*MassMercury*Period^2)/4*pi^2)^(1/3) =Radius of Orbit -- 4.55*10^6 meters?
    Then I put into ---- Tangential Velocity= (2*pi*radius)/Period --- 2.1875*10^3 m/s (too fast)?

    3. Then height above the surface of the earth is
    Height = Radius Of orbit - Radius of Mercury ---- Height = 2.117*10^3 meters

    4. The gravity at 1.50*10^6 meters away I think I have right.


    Can someone tell me if I am doing this right?
     
  7. Dec 8, 2007 #6

    Dick

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    You mean 10^6 in the height, right? Not 10^3. Otherwise, looks great!
     
  8. Dec 8, 2007 #7
    Yea, typing error
     
  9. Dec 8, 2007 #8

    D H

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    Staff Emeritus
    Science Advisor

    You're answers look good, except for some squabbling over accuracy. You expressed the mass of Mercury and the gravitational constant to three decimal places only, so any calculations done with those (i.e., questions 2, 3, and 4) are good to three decimal places or less.

    Question: Where did you get that goofy mass for Mercury? It should be 3.3022*1023 kg, or 3.30*1023 kg to three decimal places.

    Aside: Professionals in this field invaribly use a single planet-specific constant that represents the product of the gravitational constant and the planet mass, GMp[/sup] rather than the mass of the planet and G. The reason is accuracy. We know the product of G and M much more accurately than we know G or M.

    Using the published GMMercury = 22032 km3/s2, I get an orbital radius of 4573.1 km and a tangential velocity of 2.1949 km/s. In other words, comparable with your answers. You should use your answers, not mine, as you need to use the givens which include that goofy mass of Mercury.
     
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