Is Potential Energy Different for a Falling Chest at High Altitudes?

[possum30540]

Homework Statement

Imagine that a treasure chest of mass m=100kg falls to the ground from a jet flying at an altitude of 35 km.
(a) calculate the chest's initial potential energy according to formula V(z)=mgz
(b) caculate the chest's initial potential energy according to formula V(r)=-G(m1m2/r)
(c) Are answers different? Explain why this is.
(d) Caculate change in chest's ptoential energy during its fall according to formula V(z)=mgz
(e) Calculate change in chest's potential energy during its fall according to formula
V(r)=-G(m1m2/r)
(f) Are answers different? Explain why this is.

Homework Equations

V(z)=mgz

V(r)=-G(m1m2/r)

r=35000m + re = 6,415,000m

re=center of Earth to surface=6,380,000m

The Attempt at a Solution

(a) I know this answer is 3.4x10^7 J
(b)V(chest)=[-3.99x10^14 Jm/kg(100Kg)]/6,415,000m = 3.41x10^7J (WRONG)
(c) The answers should not be the same because the formula used in A and B are for different things. The formula used in A is only for kinetic energy near the Earth's surface while the formula used in B should be used for items in space (Not very clear on this)
(d) V(chest)=mg(change in z)=-3.4x10^7J (is a negative number reasonable??)
(e) V(chest)=-GMm[(1/re)-(1/re+r)]=-3.4x10^5 J
(f) The answers should not be the same because the formula used in A and B are for different things. The formula used in A is only for kinetic energy near the Earth's surface while the formula used in B should be used for items in space (Not very clear on this).

 

[anathema]

Also, the change in potential energy calculated in part (e) is much smaller than the one calculated in part (d) because the chest is falling from a much higher altitude in part (e).

 

[baba89]

Additionally, the change in potential energy is different in the two formulas because the height (z) and the distance from the center of the Earth (r) are different measurements.