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Escape Velocity and Earth's Gravity

  1. Jul 21, 2009 #1
    The Saturn V rocket for the Apollo missions had rocket reignition at stage 3 to boost speed from 28,160 km/h to 39,430km/h at a stage 2 altitude of 185km.

    I got these figures from this site: http://www.aerospaceguide.net/saturn_5.html

    I calculated Earth's escape velocity at 11.029 km/s (39,705km/h) at this altitude. Can you confirm this is correct (v= sqrt(GM/r))?

    If correct and Apollo was within ~275km/h of escape velocity (~170mph; at 99.3% of escape velocity), was there a specific reason NASA decided to get so close to that threshold, or even surpassing it would not have mattered?

    Since the Saturn V rockets started at 39,430 km/h but slowed to an average of ~5,000 km/h (~3 day trip to the moon, average distance 384,401 km / 72 hours), can you provide formulas to determine Apollo's speed at any given time during this deceleration? (For example: at 36 hours into the mission, what was the speed of the CSM/LM, assuming no other thrust movements after stage 3?)

    And did Earth's gravity continue to hold back Apollo at a constant 9.8m/s^2, or did this gravity force diminish as Apollo continued away from Earth? Can you provide formulas to determine Earth's gravity force (irrespective of lunar gravity influence, unless it makes a difference) at any given distance from the Earth? (For example: at 150,000 km from Earth, what was the pull of Earth's gravity).

    Lastly, assuming no other thrust movements after stage 3, if Apollo had initially and constantly navigated in the opposite direction from the moon, 1) how long would it have taken for Earth to pull Apollo back to Earth orbit and 2) what would have been Apollo's maximum distance just before retraction (at 0 km/h)?
     
  2. jcsd
  3. Jul 21, 2009 #2
    The formula for acceleration due to gravity (g) is:

    g = GM / (R + h)2

    G is the gravitational constant = 6.67300 * 10-11 m3 kg-1 s-2
    M is mass of Earth = 5.9742 * 1024 kg
    R is radius of Earth = 6,378,100 m
    h is height above Earth = 150,000,000 m

    That gives g = 0.016 m / s2

    This doesn't take the Moon into consideration. However, if you were to assume you were directly between the Moon and Earth you could use the same formula to determine g for the Moon, and then find the net g.
     
  4. Aug 21, 2009 #3
    should be v=sqrt(2GM/r)
     
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