# Regarding g forces on Moon and Earth

1. Jul 3, 2007

### rum2563

1. The problem statement, all variables and given/known data
For the same initial upward velocities, how many times higher will an object travel above the lunar surface (g=1.6 m/s^2) than above the surface of Earth? Assume air resistance is negligible.

2. Relevant equations
delta Y = v1y^2 X 10^2 + 1/2 X 9.8 m/s^2 X (10)^2

3. The attempt at a solution

I assumed that initial velocity should be zero for both Earth and Moon.
I specified time to be 10 seconds.

For Earth:
delta Y = 0 X 10^2 m/+ 1/2 X 9.8 m/s^2 X (100)
= 1/2 X 9.8 s^2 X (100)
= 490 m

For Moon:
delta Y = 0 X 10^2 m/+ 1/2 X 1.6 m/s^2 X (100)
= 1/2 X 1.6 s^2 X (100)
= 80 m

So am I doing this right? Because I think my final answer would be that the object would be 6 times higher above the lunar surface than the surface above the Earth.

2. Jul 3, 2007

### G01

You do not want to know the ratio of heights at 10 seconds, you want to know the ratio of heights at the highest points.

HINT:At the highest points, what are the velocities of the objects? Also, since you don't have a time value, do you have some relation you can use that doesn't involve time?

3. Jul 3, 2007

### PhanthomJay

Well, yes, the object rises 6 times higher on the moon than on the earth, but you've arrived at that answer incorrectly using an incorrect equation for delta y , which should read delta y = Vo(t) - 1/2gt^2, and Vo is not 0, it is given that Vo= Vo in both cases). But rather than get involved with time, what is the equation that relates V, y, and g?.

Last edited: Jul 3, 2007
4. Jul 4, 2007

5. Jul 4, 2007

### HallsofIvy

Staff Emeritus
Are you suggesting that gravity pulls things upward?

?? If the initial velocity of something thrown upward is 0, it won't go any distance up, either on the moon or earth!

You are answering the wrong question. What you calculated here is the distance a dropped object will fall in 10 seconds on earth or moon (taking positive y to be downward).You were asked how much farther upward an object will go on the moon rather than the earth if thrown with the same upward velocity.

Taking positive y to be upward, $v_0$ to be the initial velocity, the formulas you need are:
$$v= v_0- gt$$
$$y= v_0 t- (g/2)t^2$$

An object will go upward until v= 0 and its maximum height will be y at that t. For each value of g, set v= 0 and solve the first equation for t. Then use that t to find y.

6. Jul 4, 2007

### Andrew Mason

If you analyse the problem using energy, you will see that the maximum height is inversely proportional to g.

AM