@pagan,

Ok lets try the maths. The two relevant equations of motion i will use are the velocity and position equations.

v(t) = at + vo
where vo represents the initial velocity.

x(t) = (a.t.t)/2 + vo.t + xo
where xo is the initial position.

A dot means multiply, thus t.t means t squared

lets consider positive x as up and positive vo as up.

Lets consider positive a as up.

Set our position coordinates such that xo = 0

Further consider

**a** for the earth as

**ge** and

**a** for the moon as

**gm**. Since both are down they are negative and constant for the purposes of close approximation.

Further lets call voe as the initial upward velocity for an earth jump and vom as the intial upward velocity as a moon jump

and likewise ve te and xe for earth and vm tm and xm for moon

ve(te) = -ge.te + voe

vm(tm) = -gm.tm + vom

xe(te) = (-ge.te.te)/2 + voe.te

xm(tm) = (-gm.tm.tm)/2 + vom.tm
these are the velocity and position equations for the earth and moon.

Now lets consider a special position xh which is the high point of the jumping object as it leaves the surface with intial velocity vo and is dragged back by -g.

let th be the time when that occurs.

For such a position

**v(th) = 0** because the object is stationary at that point in the trajectory. Substitute this into the velocity equations and

Thus 0 = -ge.the +voe

and 0 = -gm.thm + vom

or

**the = voe/ge**
**thm = vom/gm**
in other words the time to reach the highest point is equal to the initial velocity needed for the jump divided by the gravitational acceleration.

Lets

**define H** as the height reached,

**which is of course x(th)**, the moment of time to reach the highest point, which we have just defined in terms of vo and g. namely th = vo/g

using the position equation therefore and substituting th for vo/g

He = (-ge.voe.voe)/(2.ge.ge) + voe.voe/ge

**He = (voe.voe)/(2.ge)**
and similarly of course

**Hm = (vom.vom)/(2.gm)**
which means that the height reached is the initial velocity of the jump squared divided by twice the gravitational acceleration.

Now lets compare equivalent jumps of height. Ie

**make Hm = He**
this means that

(voe.voe)/(2.ge) = (vom.vom)/2.gm

but we know that ge = 6.gm because the gravitational pull of the earth is 6 times stronger

So

**voe.voe = 6.vom.vom**

for a jump of the same height. We can see that you need a greater initial velocity to get the same height on the earth compared to the moon, as expected.

But now lets compare the time it takes to reach that same height. Remember that the time to reach the highest point is equal to the initial velocity needed for the jump divided by the gravitational acceleration.

the = voe/ge

thm = vom/gm

lets compare them as a ratio

**thm/the = (vom.ge)/(voe.gm)**
but again ge = 6.gm

so

**thm/the = 6.vom/voe**
but we also have for the same position

**voe.voe = 6.vom.vom**

or
**voe = vom.(square root of 6)**
thus by substitution

**thm/the = (6.vom)/(vom.(square root of 6)) = square root of 6 = 2.45**
Further, by symmetry, the return fall to the surface will take the same time as reaching the greatest height in each case and thus the proportional comparison of thm/the will remain the same.

To confirm this take the original position equation and solve for x=o (the surface)

One solution is t=0 and the other is t = 2.vo/a which is exactly twice the time to reach the highest point which was th = vo/a.

**A jump looks like time has slowed down by a factor of 2.45 when comparing the times off the surface.**
I enjoyed that!