# Homework Help: Gravitational acceleration comparison

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1. Nov 18, 2015

### Ellie Snyder

1. The problem statement, all variables and given/known data
Suppose an object of length “l” is located a distance “r” from a gravitating object of mass “M.” From physics you will learn that the gravitational acceleration is GM/r^2. Derive the difference in gravitational acceleration between distance “r” and distance “r+l” from the object. Show that as long as “l” is small compared to “r” (i.e., r >>l), the result is (2GM/r^3)l. Calculate this difference for the following two cases. What would happen to each person?

a). A person of height l=170 cm located r=1000 km from a 1.5 MSun neutron star.

b). The same person a distance 10^10 km (i.e., the width of a Solar System) from a 10^9 MSun black-hole as could be present in the nucleus of a typical galaxy.

2. Relevant equations
GM/r^2
(2GM/r^3)l

3. The attempt at a solution
I thought the difference would just be GM/(r+l)^2 - GM/r^2, and if l was small enough as compared to the other values the difference would be virtually 0. I don't know how the (2GM/r^3)l is derived and that's where I'm stuck. For the two people, would the given values be plugged into the regular gravitational acceleration equation or the one derived for the difference?

Last edited: Nov 18, 2015
2. Nov 18, 2015

### PeroK

Welcome to PF!

You need go be able to make an effort at solving the problem yourself. If necessary, by revising the course material on which the question is based. Do you really know nothing about gravity?

3. Nov 18, 2015

### Ellie Snyder

I edited to include my initial thought process.

4. Nov 18, 2015

### Staff: Mentor

Put that over a common denominator and keep solving...

5. Nov 18, 2015

### Ellie Snyder

Do you mean give the two terms a common denominator of r^2(r+l)^2, which yields (r^2-(r+l)^2)/(r^2(r+l)^2)?
Or do you mean put that entire thing over some common denominator?

6. Nov 19, 2015

### PeroK

Do you know about the Binomial expansion for negative powers? That's usually the trick in these cases where you have one variable $<<$ another. That's almost certainly what you're expected to use here.

Note that $\frac{GM}{r^2} > \frac{GM}{(r+l)^2}$ so I would set $a = \frac{GM}{r^2} - \frac{GM}{(r+l)^2}$ so you get a positive difference. Then hit this with the Binomial theorem.