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    1) A spacecraft is in a circular orbit around the moon. The radius of the orbit is 4.0 x10^6 m. The spacecraft orbits the Moon 3.81 times in 24.0hours, centripetal acceleration is 0.31ms^-2.

    The gravitational force F on the space craft is given by F = K/r^2
    where r is the radius of the prbit and K is a constant.


    i) Write down an expression for the centripetal force on a mass m moving in a circular orbit r with angular velocity w.

    I just did F=m.r.w^2


    ii) By equating the expression for gravitaional force F and answer for i) calculate the value of the radius of the orbit for which the period of the spacecraft is 12.0 hours.

    Surely you need the mass of the moon?
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    mrw² = k/r²
    m(0.31)(16x10^12) = k
    m(4.96x10^12) = k

    Now think about when the period is 12hrs. 2pi radians in 12*60*60 seconds.
    mr³w² = k
    mr³[2pi/(43200)]² = k
    As k is constant:
    mr³[2pi/(43200)]² = m(4.96x10^12)
    m =/= 0 so we can cancel it:
    r³[2pi/(43200)]² = (4.96x10^12)
    r = cube root { (4.96x10^12) / [2pi/(43200)]² }
    =6166373.28 m
    =6.17 x 10^6 m (3sf)
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    Off topic, but it seems I have the same birthday as you SsEe.

    Meh.
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    (Original post by SsEe)
    mrw² = k/r²
    m(0.31)(16x10^12) = k
    m(4.96x10^12) = k

    Now think about when the period is 12hrs. 2pi radians in 12*60*60 seconds.
    mr³w² = k
    mr³[2pi/(43200)]² = k
    As k is constant:
    mr³[2pi/(43200)]² = m(4.96x10^12)
    m =/= 0 so we can cancel it:
    r³[2pi/(43200)]² = (4.96x10^12)
    r = cube root { (4.96x10^12) / [2pi/(43200)]² }
    =6166373.28 m
    =6.17 x 10^6 m (3sf)
    aaahhh good stuff mate. Cheers!!
 
 
 

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