Lack of air resistance VS lack of gravity

[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.

 

[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.

 

[hexhunter]

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.

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.

 

[brewnog]

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

 

[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.

 

[Integral]

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.

 

[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

 

[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 I am just on the wrong track over all. any help?

 

[Integral]

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.

 

[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.

 

[Integral]

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.

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 h = R_1 - R_2 [/tex] it is the same.

 

[whozum]

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

 

[Integral]

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.

 

[Janus]

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

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.