# Gravity Meter Problem

1. Feb 18, 2007

### Gammage

1. The problem statement, all variables and given/known data
A superconducting gravity meter can measure changes in gravity of the order = 1×10-11 (delta g)/g. You are hiding behind a tree holding the meter, and your 78 kg friend approaches the tree from the other side. How close to you can your friend get before the meter detects a change in g due to his presence?

2. Relevant equations
I know that Fgrav = G (m1m2)/r^2 and G = 6.672e-11 Nm^2/kg^2

3. The attempt at a solution
My first thought was to assume the m2 was 1kg and set the Fgrav = 1e-11 N, but this was wrong.

Last edited: Feb 18, 2007
2. Feb 18, 2007

### Kurdt

Staff Emeritus
You are using the equation that describes the force between two objects due to gravity. What a gravity meter detects is variance in the gravitational field. What physics do you know of the gravitational field?

3. Feb 18, 2007

### Gammage

Very Little

I dont think we have discussed yet in class, let me check the book for any references. This is a first year Mechanics course for natural science majors.

4. Feb 18, 2007

### Gammage

Hmm, there is a very brief explaination in the chapter we are in showing that the field is equal to G(M)/r^2, which is simular to what i did before but this would give me N/kg, do i factor his mass in again to get some cancelations?

5. Feb 18, 2007

### Kurdt

Staff Emeritus
The gravitational field is basically the acceleration due to gravity and is given by

$$g =G\frac{M}{r^2}$$

where M is the mass of the Earth. In this question you will essentially need to work out how far away the 78kg (i.e. use his mass instead of the Earths) guy has to be to have an acceleration toward him due to his own gravity of 1x10-11 ms-2.

6. Feb 18, 2007

### Gammage

the 1x10^-11 is mentioned in the problem as delta g divided by g. when I set g = 1x10^-11 and M = 78kg, and solved for r the answer is wrong.