Energy to move mass (potential energy)

[merced]

Homework Statement

How much energy is required to move a 900 kg mass from the Earth's surface to an altitude 3 times the Earth's radius?

Homework Equations

\Delta U = GMm[\frac{1}{R_1} - \frac{1}{R_2}]
G = 6.67 x 10^{-11} Nm^2/kg^2
R_1 = R_E = 6.37 x 10^6 m
M = 5.98 x 10^{24} kg

The Attempt at a Solution

I plugged in everything with R_1 = radius of earth
and R_2 = 3(radius of earth) + 1 radius of Earth = 4 radius of earth...and got \Delta U = 4.23 x 10^{10}.

The answer is 3.76 x 10^{10}, where R_2 = 3(radius of earth).

Why don't you use the radius of the Earth plus the altitude (i.e. 4x radius of earth)? For the first potential energy, I use the radius as the radius of the earth, so I don't see why you use only the altitude for the second potential energy.

 

[nrqed]

Homework Statement

How much energy is required to move a 900 kg mass from the Earth's surface to an altitude 3 times the Earth's radius?

Homework Equations

\Delta U = GMm[\frac{1}{R_1} - \frac{1}{R_2}]
G = 6.67 x 10^{-11} Nm^2/kg^2
R_1 = R_E = 6.37 x 10^6 m
M = 5.98 x 10^{24} kg

The Attempt at a Solution

I plugged in everything with R_1 = radius of earth
and R_2 = 3(radius of earth) + 1 radius of Earth = 4 radius of earth...and got \Delta U = 4.23 x 10^{10}.

The answer is 3.76 x 10^{10}, where R_2 = 3(radius of earth).

Why don't you use the radius of the Earth plus the altitude (i.e. 4x radius of earth)? For the first potential energy, I use the radius as the radius of the earth, so I don't see why you use only the altitude for the second potential energy.

Your reasoning is entirely correct, r_f {\em should} be 4 R_E. If they used three times the radius of the Earth, they made a mistake, since "altitude" is defined to be measured above ground.

 

[merced]

Ok, If I only can tell my professor before the next test!

Thanks!