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.

Short answer

(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 F_T<W.

Step-by-step solution

 Step 1: Given Data

• Thrust force = 30000 N.
• Mass of the module=10000 kg.

(a) Determine the magnitude of acceleration during takeoff from the Moon

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,F_Tis the thrust force, g_Moon 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 g_Moon, and 30000 N forF_T, 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.

Step 3:(b) Determine whether the module would take off from Earth

Calculate the weight of the module on Earth,

\(W = m{g_{{\rm{Earth}}}}\)

Here, W is the weight of the module, and g_Earth 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.