A lunar module weighs 16 tons on the surface of Earth. How much work is done in propelling the module from the surface of the moon to a height of 50 miles? Consider the radius of The moon to be 1100 miles and its force of gravity to be one-sixth that of Earth.
A) $1.35 \times 10 ^ { 12 } \mathrm { ft } - \mathrm { lb }$
B) $1.35 \times 10 ^ { 9 } \mathrm { ft } - \mathrm { lb }$
C) $6.73 \times 10 ^ { 8 } \mathrm { ft } - \mathrm { lb }$
D) $4.09 \times 10 ^ { 9 } \mathrm { ft } - \mathrm { lb }$
E) $4.09 \times 10 ^ { 12 } \mathrm { ft } - \mathrm { lb }$

Question

A lunar module weighs 16 tons on the surface of Earth. How much work is done in propelling the module from the surface of the moon to a height of 50 miles? Consider the radius of The moon to be 1100 miles and its force of gravity to be one-sixth that of Earth. 
A) $1.35 \times 10 ^ { 12 } \mathrm { ft } - \mathrm { lb }$
B) $1.35 \times 10 ^ { 9 } \mathrm { ft } - \mathrm { lb }$
C) $6.73 \times 10 ^ { 8 } \mathrm { ft } - \mathrm { lb }$
D) $4.09 \times 10 ^ { 9 } \mathrm { ft } - \mathrm { lb }$
E) $4.09 \times 10 ^ { 12 } \mathrm { ft } - \mathrm { lb }$

Quizplus · Accepted Answer

First, we need to find the mass of the lunar module in kilograms since the standard unit for work is joules.

$16$ tons $= 32,000$ pounds $= 14,515.4$ kilograms

The potential energy of an object is given by the formula $PE = mgh$, where $m$ is the mass of the object, $g$ is the force of gravity acting on the object and $h$ is the height the object is lifted to.

The force of gravity on the moon is one-sixth that of Earth, so $g = \frac{1}{6} 9.81 \frac{m}{s^2} = 1.635 \frac{m}{s^2}$

The distance from the surface of the moon to a height of 50 miles is:

$50\mathrm{mi} = 50 \cdot 5280\mathrm{ft/mi} = 264,000\mathrm{ft}$

The radius of the moon is given as 1100 miles, or 5,808,000 feet. We need to add this to the height of 50 miles to obtain the distance the lunar module is from the center of the moon:

$264,000\mathrm{ft} + 5,808,000\mathrm{ft} = 6,072,000\mathrm{ft}$

We can now find the potential energy gained by the module as it travels from the surface of the moon to a height of 50 miles:

$PE = mgh = 14,515.4\mathrm{kg} \cdot 1.635 \frac{m}{s^2} \cdot 6,072,000\mathrm{ft} $

Converting to joules:

$PE = 1.35 	imes 10^9\mathrm{J}$

Finally, we need to convert joules to foot-pounds since that is the specified unit in the answer choices:

$1\mathrm{J} = 0.737562\mathrm{ft-lb}$

$\implies PE = 9.96 	imes 10^8 \mathrm{ft-lb} \approx 1.35 	imes 10^9 \mathrm{ft-lb}$

Therefore, the answer is B.

A Lunar Module Weighs 16 Tons on the Surface of Earth