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Posted by on Sunday, December 16, 2007 at 8:40am.

Consider a spaceship located on the Earth-Moon center line (i.e. a line that intersects the centers of both bodies) such that, at that point, the tugs on the spaceship from each celestial body exactly cancel, leaving the craft literally weightless. Take the distance between the centers of the Earth and Moon to be 3.90E+5 km and the Moon-to-Earth mass ratio to be 1.200E-2. What is the spaceship's distance from the center of the Moon?

I'm still a little confused on what I have to do for this one. Which distance equation do I need to use??

  • Physics - , Sunday, December 16, 2007 at 8:41am

    Oh wait...do I need to use F = GM1M2/d^2?

  • Physics - , Sunday, December 16, 2007 at 8:53am

    What you need to do is balance the forces.

    forceshipEarth= forceshipmoon
    GmMe/d^2=GmMm/d2^2

    Mm/Me= (d2/d)^2

    where mm is mass moon, me mass earth, d2 is distance from craft to moon, and d is the distance from craft to earth.

    You know d+d2=3.9E5km, so solve for d2.

    I would start with taking the square root of each side of the equation above, then solving for d2.

  • Physics - , Sunday, December 16, 2007 at 9:14am

    Ok I understand that. But what is the distance from craft to earth?

  • Physics - , Sunday, December 16, 2007 at 9:58am

    << At what distance from the earth toward the moon will the gravitational force of attraction towards the moon be equal and opposite to the gravitational force of attraction towards the earth? >>

    The Law of Universal Gravitation states that each particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Expressed mathematically,
    F = GM(m)/r^2
    where F is the force with which either of the particles attracts the other, M and m are the masses of two particles separated by a distance r, and G is the Universal Gravitational Constant. The product of G and, lets say, the mass of the earth, is sometimes referred to as GM or mu (the greek letter pronounced meuw as opposed to meow), the earth's gravitational constant. Thus the force of attraction exerted by the earth on any particle within, on the surface of, or above, is F = 1.40766x10^16 ft^3/sec^2(m)/r^2 where m is the mass of the object being attracted and r is the distance from the center of the earth to the mass.
    The gravitational constant for the earth, GM(E), is 1.40766x10^16ft^3/sec^2. The gravitational constant for the moon, GM(M), is 1.7313x10^14ft^3/sec^2. Using the average distance between the earth and moon of 239,000 miles, let the distance from the moon, to the point between the earth and moon, where the gravitational pull on a 32,200 lb. satellite is the same, be X, and the distance from the earth to this point be (239,000 - X). Therefore, the gravitational force is F = GMm/r^2 where r = X for the moon distance and r = (239000 - X) for the earth distance, and m is the mass of the satellite. At the point where the forces are equal, 1.40766x10^16(m)/(239000-X)^2 = 1.7313x10^14(m)/X^2. The m's cancel out and you are left with 81.30653X^2 = (239000 - X)^2 which results in 80.30653X^2 + 478000X - 5.7121x10^10 = 0. From the quadratic equation, you get X = 23,859 miles, roughly one tenth the distance between the two bodies from the moon. So the distance from the earth is ~215,140 miles.
    Checking the gravitational pull on the 32,200 lb. satellite, whose mass m = 1000 lb.sec.^2/ft.^4. The pull of the earth is F = 1.40766x10^16(1000)/(215,140x5280)^2 = 10.91 lb. The pull of the moon is F = 1.7313x10^14(1000)/(23858x5280)^2 = 10.91 lb.
    This point is sometimes referred to as L2. There is an L5 Society which supports building a space station at this point between the earth and moon. There are five such points in space, L1 through L5, at which a small body can remain in a stable orbit with two very massive bodies. The points are called Lagrangian Points and are the rare cases where the relative motions of three bodies can be computed exactly. In the case of a body orbiting a much larger body, such as the moon about the earth, the first stable point is L1 and lies on the moon's orbit, diametrically opposite the earth. The L2 and L3 points are both on the moon-earth line, one closer to the earth than the moon and the other farther away. The remaining L4 and L5 points are located on the moon's orbit such that each forms an equilateral triangle with the earth and moon.

    I hope this is what you were looking for.

  • Physics - , Sunday, December 16, 2007 at 12:20pm

    Actually, that just made me even more confused. I still don't know what I'm supposed to use as the distance from the spaceship to the earth.
    Thanks, tho.

  • Physics - , Sunday, December 16, 2007 at 7:17pm

    From the quadratic equation, you get X = 23,859 miles, roughly one tenth the distance between the two bodies from the moon.

    So the distance from the earth is ~215,140 miles.

    Checking the gravitational pull on the 32,200 lb. satellite, whose mass m = 1000 lb.sec.^2/ft.^4. The pull of the earth is F = 1.40766x10^16(1000)/(215,140x5280)^2 = 10.91 lb. The pull of the moon is F = 1.7313x10^14(1000)/(23858x5280)^2 = 10.91 lb.

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