# Is this the approximate gravity of Mars?

1. Feb 24, 2009

### .:Endeavour:.

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:
$$\frac{45.359 kg}{444.518 N} = \frac{17.237 kg}{X}$$

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

7662.156766 = 45.359X

$$\frac{7662.156766}{45.359} = X$$

168.9225 N = X
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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. Feb 25, 2009

### Nabeshin

Last edited by a moderator: Apr 24, 2017
3. Feb 25, 2009

### DaleSwanson

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 = $$\frac{(6.67300 * 10^{-11}) * (6.4191 * 10^{23})}{3,397,000^{2}}$$

g = $$\frac{42834654300000}{11539609000000}$$

g = 3.71