Is this the approximate gravity of Mars?

[.: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 F_g of Mars, I did this kind of equation:
<br /> \frac{45.359 kg}{444.518 N} = \frac{17.237 kg}{X}<br />

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

7662.156766 = 45.359X

<br /> \frac{7662.156766}{45.359} = X<br />

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

F_gravity = 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].

 

[Nabeshin]

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 F_g of Mars, I did this kind of equation:
<br /> \frac{45.359 kg}{444.518 N} = \frac{17.237 kg}{X}<br />

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

7662.156766 = 45.359X

<br /> \frac{7662.156766}{45.359} = X<br />

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

F_gravity = 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].[/QUOTE]

I don't understand what you're trying to do. You're trying to find the gravitational field strength at the surface of mars? Is that what you mean by F_g?

If so, you are given: F_gMars=.38*F_gEarth Which solve the problem for you considering you know the field strength of Earth is 9.8N/kg...

 

[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

Google’s calculator is great.
http://www.google.com/search?q=(G+*+mass+of+mars)+%2F+radius+of+mars+^+2

 

[.:Endeavour:.]

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 F_g. Because mass is constant and doesn't changes with the gravitational pull, but only weight does change, right? Thank you for your corrections.