What is the relationship between the value of g and the distance from Earth?

[becca4]

Homework Statement

Homework Equations

So I know that there's an inverse relationship between the value of g and r; the farther I get away from earth, the weaker the gravitational force -> weaker acceleration...

BUT I'm not sure how the (1-x)^-2 thing is related, need help with that. Also, how do I set up for Newton's universal law of gravitation?

The Attempt at a Solution

Uh... In process. )

Thanks in advance!

 

[andrevdh]

From Newton's universal gravitational law we have:

$$W = mg = GMm\ \frac{1}{r^2}$$

following

$$g = GM\ \frac{1}{r^2}$$

now use the hint ...

 

[becca4]

I still don't really understand... :grumpy:

 

[esalihm]

in Newtons equation, put distance=(re+delta r)

then write it as d= re(1+ (delta r/re))

d has a power of -2

and at 100 km. "delta r/re" is between -1 and 1 since the radius of the Earth is greater than 100 km

does this help or do you need more?

 

[becca4]

I think I almost have it. The only thing that I'm still wondering is about the relation part, what does the minus sign mean?

 

[esalihm]

because you are given an equation that is stating the change in g (delta g), that minus indicates a decrease.
think about what happens to g as the distance from the center of the Earth increases by considering Newton's equation

 

[andrevdh]

the minus sign means that for a positive $$\Delta r$$ (going further up) the change in the gravitational acceleration decreases (is negative). So that the new gravitational acceleration is given by

$$g_{new} = g_{old} + \Delta g$$

 

[becca4]

Ok, sorry to bring this up again.

Talking to some of my classmates, some of them think that this is supposed to be a proof using Taylor series stuff, and that at the end that's where the -2 comes from. Any thoughts on that?