91Ó°ÊÓ

• 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,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.

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.