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Is this the approximate gravity of Mars?

  1. Feb 24, 2009 #1
    I've found this site that states: "The surface gravity on Mars is only about 38% of the surface gravity on Earth." So a 100 lbs object will weigh about 38 lbs or a 45.359 kg will be on Mars 17.237 kg (converting 38 lbs in kg). So to find the Fg of Mars, I did this kind of equation:
    [tex]
    \frac{45.359 kg}{444.518 N} = \frac{17.237 kg}{X}
    [/tex]

    (444.518)(17.237) = (45.359)(X)

    7662.156766 = 45.359X

    [tex]
    \frac{7662.156766}{45.359} = X
    [/tex]

    168.9225 N = X
    -----------------------------------
    Now that we know the weight of the object in newtons on Mars' surface, then:

    17.237 kg ÷ X = 168.9225 N

    X = 0.10204876 N

    Fgravity = 0.10204876 N

    So the force of gravity on Mars surface is approximately 0.10204876 Newtons, right? The site that I found the information at is: [URL]http://coolcosmos.ipac.caltech.edu/cosmic_kids/AskKids/mars_gravity.shtml[/ur].
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Feb 25, 2009 #2

    Nabeshin

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    Last edited by a moderator: Apr 24, 2017
  4. Feb 25, 2009 #3
    The formula for finding g is:
    g = GM / R2

    Where G is the gravitational constant, M is the mass of the object (Mars), and R is the radius of the object. For Mars this would be:

    g = [tex]\frac{(6.67300 * 10^{-11}) * (6.4191 * 10^{23})}{3,397,000^{2}}[/tex]

    g = [tex]\frac{42834654300000}{11539609000000}[/tex]

    g = 3.71

    Google’s calculator is great.
    http://www.google.com/search?q=(G+*+mass+of+mars)+%2F+radius+of+mars+^+2
     
  5. Feb 25, 2009 #4
    Thank you for your help. It started to look wrong like I did it now that I'm looking it over. It was 17.237 kg that would trow everything off where you have to find what is 0.38 on Mars' surface times its Fg. Because mass is constant and doesn't changes with the gravitational pull, but only weight does change, right? Thank you for your corrections.
     
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