Answer:
Explanation:
You need something that relates distance to what the gravitational pull is. You can set up a complex sort of proportion. What you need is a number that is comparable to 9.81 or you can just use the Gravitational Force formula with a 4 tier fraction.
Givens
x = the additional distance toward outer space above the radius of the earth.
G is the gravitational constant.
m1 = the person's mass (which does not change no matter where you are).
m2 = the earth's mass
F1 = 588 N
F2 = 300 N
Formula
[tex]\frac{F1}{F2} = \frac{588 N}{300N}=\frac{\frac{Gm1*m2/}{6400^2} }{\frac{G*m1*m2}{(6400 + x)^2} }[/tex]
Solution
G*m1*m2 all cancel. So what you get looks like this.
[tex]\frac{588}{300} = \frac{(6400 + x)^2}{6400^2}[/tex]
Cross Multiply
588 * 6400^2 = 300*(6400+x)^2 Now all you need do is solve for x.
x will be in km.
588*40960000 = 300 * (40960000 + 12800x + x^2)
1.2288*10^10 + 3840000x + 300x^2 = 2.408448*10^10
300x^2 + 3840000x + 1.2288*10^10 = 2.408448 * 10^10
Subtract 2.409448 * 10^10 from both sides.
300x^2 + 3840000x - 1,179648 * 10^10
Now use the quadratic formula
I'm guessing I should have converted this to meters because I'm getting ridiculous numbers. They are already large enough as you can see. The method is correct, even if the numbers are not.
