# Error Propogation - E of Gravity due to r

1. Sep 16, 2007

### skateza

1. The problem statement, all variables and given/known data
The acceleration due to gravity is given by Newton's universal law of gravitation as:
g=(GM)/r^2

Derive an algebraic expression for the error in gravity (Eg) due to the uncertainty in r. You may assume that the errors in G & M = 0; (EG & EM = 0)

Let E=error in for internet purposes
2. Relevant equations

g=(GM)/r^2

3. The attempt at a solution

r^2=(GM)/g
r=(GM/g)^(1/2)
r=[(GM)^(1/2)]/[g^1/2]

This is calculus, i can't see a classical algebraic way of doing this..

any help plz?

2. Sep 16, 2007

### genneth

Suppose f is a function of r. Then:

$$f + \delta f \approx f(r) + \delta r f^{\prime}(r)$$

so $$\delta f \approx \delta r f^{\prime}(r)$$

Can you see why this is the case?

3. Sep 16, 2007

### skateza

in the second half of that you are taking the derivative of the function, as far as i am convinced this course is an algebra based course and i'm trying to figure out an algebraic response to this question, without taking the derivative at any point.... would it be
r^2+Er^2=GM/g+Eg

4. Sep 16, 2007

### genneth

Okay, then:

$$g + \delta g = \frac{GM}{(r + \delta r)^2}$$

Knowing that $$g = \frac{GM}{r^2}$$

Apply binomial expansions as necessary.