14PE
Question
Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 10,000 kg. The thrust of its engines is 30,000 N.
(a) Calculate its the magnitude of acceleration in a vertical takeoff from the Moon.
(b) Could it lift off from Earth? If not, why not? If it could, calculate the magnitude of its acceleration.
Step-by-Step Solution
Verified(a) The magnitude of the acceleration is 1.33 m/s2.
(b) The module will not take off from Earth because of the value of FT<W.
- Thrust force = 30000 N.
- Mass of the module=10000 kg.
Apply Newton’s second law of motion,
\(\begin{array}{c}{F_T} - w = ma\\a = \frac{{{F_T} - w}}{m}\\a = \frac{{{F_T} - m{g_{Moon}}}}{m}\\a = \frac{{{F_T}}}{m} - {g_{Moon}}\end{array}\)
Here, FT is the thrust force, gMoon is the acceleration due to gravity on Moon, m is the mass of the module, and a is the acceleration.
Substitute 10000 kg for m and 1.67 m/s2 for gMoon, and 30000 N for FT, and we get,
\(\begin{array}{c}a = \frac{{30000\;{\rm{N}}}}{{10000\;{\rm{kg}}}} - 1.67\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\ = \frac{{30000\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}{{{\rm{10000}}\;{\rm{kg}}}} - 1.67\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\ = \left( {3 - 1.67} \right)\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\ = 1.33\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{array}\)
Hence, the magnitude of the acceleration is 1.33 m/s2.
Calculate the weight of the module on Earth,
\(W = m{g_{{\rm{Earth}}}}\)
Here, W is the weight of the module, and gEarth is the acceleration due to gravity on Earth.
Substitute 10000 kg for m and 9.8 m/s2 for g in the above equation, and we get,
\(\begin{array}{c}W = 10000\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\ = 98000\;{\rm{N}}\end{array}\)
Since the value of FT is less than that of W; hence, the module will not take off from the Earth.