Calculating Vertical Jump on Earth and the Moon

[alison16]

Homework Statement

On Earth, an average person's vertical jump is 0.40m. What is it on the Moon?

Homework Equations

Fg 1 on 2 = G(m1m2/r^2)
G= 6.67 x 10^-11

The Attempt at a Solution

r= 0.40m
F Earth on person on Earth's surface = 6.67x10^-11(m Earth x m person)/(0.40)^2
r=?
G moon = ?
F moon on person on Moon's surface = G(m moon x m person)/r^2[/B]
I'm pretty stuck, and I don't understand if Earth is helpful in solving it or not. The textbook does not give any relevant values/equations except for the ones I typed under 1 and 2. Is G of the moon also 6.67 x 10^-11?

 

[gneill]

Hi Alison16, Welcome to Physics Forums.

G is the same for Earth and the Moon and anywhere else; it's what's called a Universal Constant and applies everywhere.

What information do you have about the masses or relative masses of the Earth and Moon? Or perhaps you were given some information about the relative strength of gravity on the Moon versus Earth? You might have to look these things up in your text or course notes.

 

[alison16]

Thanks! I think I found more information. In an example problem, it uses 6.0 x 10^24 kg as Earth's mass. And Moon's as 7.35 x 10^22 kg. Will I use Earth's mass to find the m person? And then use that information to solve F moon on person on Moon's surface = G(m moon x m person)/r^2?

 

[gneill]

Thanks! I think I found more information. In an example problem, it uses 6.0 x 10^24 kg as Earth's mass. And Moon's as 7.35 x 10^22 kg. Will I use Earth's mass to find the m person? And then use that information to solve F moon on person on Moon's surface = G(m moon x m person)/r^2?

I think a better approach would be to find the acceleration due to gravity on the Moon's surface (you should already know what it is at the Earth's surface), and then think about conservation of energy. When a person makes a jump, they give themselves some initial kinetic energy depending upon their strength. Assume that it's the same amount of energy they can manage in both cases.

 

[HallsofIvy]

The difference will depend upon two things- the difference in masses of the Earth and moon and the difference in radii of the eath and moon.
For a person of mass m on the Earth $F_e= \frac{GmM_e}{r_e^2}$ where $M_e$ the mass of the Earth and $r_e$ is its radius. Similarly for a person of mass m on the Earth $F_m= \frac{GmM_m}{r_m^2}$ where $M_m$ the mass of the moon and $r_m$ is its radius.
The ratio of force on the moon to force on the Earth is one divided by the other $\frac{GmM_m}{r_m^2}\frac{r_e^2}{GmM_e}$. The "Gm" terms cancel leaving $\frac{M_m r_e^2}{M_er_m^2}= \frac{M_m}{M_e}\left(\frac{r_e}{r_m}\right)^2$.
Now, how is the height an object can rise to determined by the gravitational force on it?

 

[alison16]

I understand now! Thank you all very much! :)