# Need Help Gravitational Problem

1. Nov 14, 2004

### Dragoon

Need Help!!!! Gravitational Problem

I am kinda lost here was hoping someone could help me out here
thanks a ton

The year is 2115. There is a high jump competition on planet Mars. An athlete of mass 72 kg who has been clearing a height of 2.44 m on Earth just the week before is competing here. What is the height this athlete can expect to clear in this competition?
Relevant data:
Mass of Mars = 6.42·1023 kg
Mass of Earth = 6.00·1024 kg
Radius of Mars = 3.40·103 km
Radius of Earth = 6.38·103 km

2. Nov 14, 2004

### Dragoon

i dont really have any equations that deal with this problem except
Force of gravit=Gm1m2/r2
i dont know how to do it

3. Nov 14, 2004

### da_willem

By equating F=ma and Newtons law of gravitation you can find the gravitational acceleration on the planets.

4. Nov 14, 2004

### Dragoon

ok but how would i do that for the moon all i have its mass but i need a force or acceleration and since i want excel how would i derive a force?

5. Nov 14, 2004

### tony873004

Typing "Surface gravity of Mars" into Yahoo search engine gives me

http://blueox.uoregon.edu/~jimbrau/astr121/Notes/Mars/marsgrav.html

as the first result. The bottom of the page gives you a very simple formula for computing the surface gravity of Mars as a percentage of Earth's. All you need now is the Earth's surface gravity which is a number you should have memorize already.

After getting Mars' surface gravity expressed as an acceleration, your problem just becomes a regular kinmatics problem.

Last edited by a moderator: Apr 21, 2017
6. Nov 14, 2004

### Janus

Staff Emeritus
Nitpick here: To accurately solve this problem in reality, not enough info is given. When a high jumper clears a bar of 2.44 meters on the Earth, he is not actually lifting his mass that far. What he is doing is getting his center of Mass over the bar. Since his center of mass is not at ground level but somewhere above that, the distance he lifts his mass against gravity is equal to the height he clears minus the height of his center of gravity.

So this high jumper would not clear 4.88 meters on a world with 1/2 the gravity of Earth. He would only be able to clear twice the difference between 2.44 meters and his CoG plus the height of his CoG. For instance if his CoG was 1 m off the ground he would clear 2*1.44 + 1 = 3.88 meters.

Since this problem never gives the height of the jumper's CoG, you can't arrive at a "correct" answer.

Last edited: Nov 14, 2004
7. Nov 15, 2004

### Dragoon

Thats what i thought Janus

It seemed to me that there was information missing but since my class is a computer class the computer wouldnt accept an explanation such as that but this is what the computer said i was supposed to do after i got it wrong.

The correct answer is: 6.47 m.
The work the athlete can do is a function of the conditioning of the athlete and won't change from one week to the other. This work is:

W = Fh (= force · height) = mMh/R2
Writing this equation for Earth and Mars, and realizing that the work is the same, we get:
mMehe/Re2 = mMmhm/Rm2
=> hm = he(Me/Mm)(Rm/Re)2
= 6.47 m.