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Newton's Law of Gravitation Feynman Lectures

  1. Aug 22, 2011 #1
    Hi all,
    I've been reading the Feynman Lectures on Physics and I've stumbled on something. I understand the theory but not how they arrived at the answer. It's to do with firing a bullet from a gun and working out the speed it would need to travel in a curve around the earths surface in order to be at the same height that it started out. They prove it using plane geometry which has confused me because as far as I understand it the bullet would need to travel in an arc for which ultimately the equation will need an angle. The answer is about 5 miles a second. Please can someone explain to me how they would approach this problem assuming that the raduis of the earth is 4000miles and the an object will fall 16ft/sec under the influence of gravity?

  2. jcsd
  3. Aug 22, 2011 #2
    If you accelerate a projectile to the appropriate tangential velocity, it will possess insufficient velocity to escape the gravitational field just as it will be a velocity too great to allow the projectile’s descent to the planet’s surface. The projectile’s tangential velocity per second is just sufficient to counter the projectile’s rate of vertical free-fall descent per second per the rate of gravitational attraction at the given radius thereby allowing the projectile to maintain an indefinite circular orbit (assuming no frictional losses).

    The following yields the velocity (v) required to orbit per a given radius (r) and a given heavenly body’s Mass (M), with ‘G’ being the Gravitational constant, 6.67e-11:

    v = sqrt(GM / r)

    sqrt(6.67e-11 * 5.976e+24 kg) / 6,377,569.11 meters) = 7,905.7 m/s

    Or, you could apply Newton’s Centripetal Force. The earth’s gravitational acceleration (a) at its surface is 9.8 m/s^2. Applying the same earth radius (r) of 6,377,569.11 meters, we derive a required tangential velocity (v) of:

    v^2 = ar

    Hence, the required tangential orbital velocity (v) is,

    v = sqrt(ar)

    sqrt(9.8 m/s^2 * 6,377,569.11 meters) = 7,905.7 m/s

    I applied precision numerical values to demonstrate the interchangeable precision of the two equations. I hope you found this helpful.
  4. Aug 23, 2011 #3
    Hi Gnosis
    Thanks for the response. Only just had the chance to look at it.
  5. Aug 27, 2011 #4
    It was my pleasure, Mark.
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