What Physics Would Allow a Cow to Jump Over the Moon?

[diverse]

A. How much energy must a 450.Kg cow expend in jumping over the moon?(excluding the moon's gravity and the Earth's atmosphere),

B. Assuming all the work is done during the jump itself, what would her initial velocity have to be?

C. What would her final velocity be upon her return to Earth?

D. Again assuming all the work would be done during the initial jump, i.e. for a distance approximately the length of her legs ( 1 meter ), what would her acceleration need to be.

Homework Equations

EL = G Me m(1/r0 - 1/r1)

The Attempt at a Solution

Substituting values in:
G = 6.67 x 10^-11 Nm^2 / kg^2
Me = 5.98 x 10^24 kg
m = 450. kg
r0 = 6.38 x 10^6 m
r1 = 3.85 x 10^8 m

Using all those in the big equation I get this:
2.77 x 10^10 J This would be her kinetic energy for A hopefully.

For part B I just used KE = 1/2 mv^2
Substituting everything in and solving for V I get 5550 m/s

I don't quite get question C, how would she even return to Earth with that velocity?

For part D I haven't quite worked it out yet but I'am going to use
v^2 = V0^2 + 2a Delta t and just solve for the acceleration. But this assumes my velocity is right, which i'am not sure of.

Have I missed anything important in my thinking here? I can't really check my answers for reasonability

 

[denverdoc]

are ro and r1 the raddii from Earth's center to surface and the additional distance to the far side of the moon. If so those numbers look off.

I'm guessing the distance to the farside is something like 250,000mi plus another 2160 for the moons diam. that's about 250,000/.6 Km. you may have right numbers, not sure, but the moon energy term should be expressed as r0+r1 where r1=distance from surface of Earth to far side.

Elsie would get back the same way she got there, and will have the same velocity profile on the way down as up, just as a rock hurled upwards. (ignoring the obvious orbital mechanics issues)

another way to calulate the acceleration which avoids messing with time, uses V^2=2a*x where x = 1 meter.

 

[m2003]

A. To calculate the amount of energy a 450 kg cow must expend in jumping over the moon, we need to know the height of the jump. Without this information, it is not possible to accurately calculate the energy expenditure.

B. Assuming all the work is done during the jump itself, the cow's initial velocity would depend on the height of the jump. If we assume a height of 1 meter, the initial velocity would be approximately 14 m/s.

C. It is not possible for the cow to return to Earth with a velocity of 5550 m/s. This would require a significant amount of additional energy and force, which is not realistic for a cow to produce.

D. To calculate the acceleration of the cow during the jump, we would need to know the time it takes for the cow to reach the peak of the jump. Without this information, it is not possible to accurately calculate the acceleration.