# Lack of air resistance VS lack of gravity

1. Apr 2, 2005

### hexhunter

a small question, on the moon, the small amount of gravity prevents sudden falling like on earth, however would the lack of gases allow somebody jumping on the moon a much faster falling speed than on earth, because they do not have resistance from the electrons in usual earthly gases.

2. Apr 2, 2005

### brewnog

I don't know what you mean by "resistance from the electrons" but yes, if you jumped off a really tall building on the moon, you'd end up going a fair bit faster. Because there wouldn't be any forces opposing your falling motion, you would simply carry on accelerating (until splat-down) rather than achieving a terminal velocity as on earth.

3. Apr 2, 2005

### hexhunter

the negetive EM charge of an electron means that when it meets another electron they repel each other, just like magnets.

so you mean that in long distances, yes, but you would still have the slow-mo effect when jumping short distances.

4. Apr 2, 2005

### brewnog

Sorry, I was talking on the macroscopic scale (people jumping etc).

5. Apr 2, 2005

### whozum

It depends on how high you jump. Without air resistance there is no terminal velocity on the moon. However on earth, a person typically has a terminal velocity of about 150mph on earth.

6. Apr 2, 2005

### Integral

Staff Emeritus
An interesting question for someone to figure out. (I may in a bit)!

How far must you fall on the moon to reach and exceed the terminal velocity of a fall on earth? (Assume terminal velocity of 50 m/s (nice round number which is close)

I'll give some else a chance at it. Then, in a bit, when I get some time I will post a solution.

7. Apr 2, 2005

### whozum

You want the height of fallin for an impact velocity of 50m/s

v = a t

50 = (3.4)(t)

t = 14.705s

Constant acceleration of 3.4m/s^2 for 14.705s without air resistance gives:

$$\Delta X = \frac{at^2}{2} = \frac{(3.4)(14.7)^2}{2} = 367.35m$$ high. Anything above this will breach 50m/s

8. Apr 2, 2005

### whozum

I tried this using energy conservation:

2GM/R = V^2

$$R = 2GM/V^2 = (2*6.67*7.36*10^(22-11))/2500 = (98.18x10^11)/2500 = 3.927 x 10^9$$ from the moon's center (1737400m radius)

Thats MUCH bigger, but I think im just on the wrong track over all. any help?

9. Apr 2, 2005

### Integral

Staff Emeritus
According to my CRC Handbook of Chem and physics $g_m = 1.62 \frac m s$

As you did I started with

v= at

or $$t = \frac v a$$

also

$$x = \frac 1 2 a t^2$$

so
$$x = \frac 1 2 a ( \frac v a )^2$$

$$x = \frac {v^2} {2 a}$$

let v = 50 $\frac m s$ and a = 1.62 $\frac m {s^2}$

x ~ 770m

From energy considerations we have

$$mgh = \frac 1 2 m v^2$$

or

$$h = \frac {v^2} {2 g}$$

This is the same as my final expression above.

Last edited: Apr 2, 2005
10. Apr 2, 2005

### whozum

Stupid me, I was doing my problem for mars for some reason.

Is it wrong to use energy conservation in this way:

$$KE = -GPE$$

$$\frac{mv^2}{2} = -\frac{GMm}{R}$$

The kinetic energy gained will equal the loss in gravitational potential energy.

11. Apr 2, 2005

### Integral

Staff Emeritus
The potential energy is the CHANGE in height. so you would need to have a term like

$$R_1 - R_2$$

Where $R_1$ is the starting point and $R_2$ the ending. But if you just let [itex] h = R_1 - R_2 [/tex] it is the same.

Last edited: Apr 2, 2005
12. Apr 2, 2005

### whozum

Im trying to accomodate this but I can't get it. Does GMm/R simplify to mgh?

13. Apr 2, 2005

### Integral

Staff Emeritus
My apologies I should have caught this right off the bat.

$$G \frac {mM} {r^2}$$

is not energy that is gravitational FORCE. No, that expression will not get you what you want.

14. Apr 3, 2005

### Janus

Staff Emeritus
Okay,

in order to get what you want you need to do this.

The potential energy at the moon's surface would be

$$P_em =-\frac{GMm}{R_m}$$

At height h above the surface:

$$P_eh =-\frac{GMm}{R_m+h}$$
[/tex]

We want the difference so we get:

$$\Delta P_e = \frac{GMm}{R_m}-\frac{GMm}{R_m+h}$$

This is what will equal the kinetic energy of the falling mass. thus:

$$\frac{GMm}{R_m}-\frac{GMm}{R_m+h}= \frac{mv^2}{2}$$

$$GMm \left( \frac{1}{R_m}-\frac{1}{R_m+h} \right) = \frac{mv^2}{2}$$

$$\frac{1}{R_m}-\frac{1}{R_m+h}= \frac{mv^2}{2GMm}$$

$$\frac{1}{R_m}-\frac{1}{R_m+h}= \frac{v^2}{2GM}$$

$$\frac{1}{R_m}-\frac{v^2}{2GM}= \frac{1}{R_m+h}$$

$$\frac{1}{\frac{1}{R_m}-\frac{v^2}{2GM}}= R_m+h$$

$$\frac{1}{\frac{1}{R_m}-\frac{v^2}{2GM}}-R_m= h$$

If R is very large as compared to h, then this answer comes out very close to that you would get using mgh.