Question

Determine the work need to boost a satellite of mass\(1000kg\)out of the gravitational field.

Short answer

The work needed to boost a satellite of mass\(1000kg\)out of the gravitational field is\(9.83 \times {10^6}J\).

Step-by-step solution

Explanation from exercise 21.

From exercise 21, the gravitational force with which two masses\(m1\)and\(m2\)attracts each other is given by\(F = G\frac{{{m_1}{m_2}}}{{{r^2}}}\).

As the satellite is at a height of \(1000km\)after it is propelled. Thus, the distance between two bodies varies from\(R\)to\(R + 1000\).

The work done by the gravitational field to propel the satellite is computed as

\(\begin{aligned}{}W &= \int_a^b F dx\\ &= \int_R^{R + 1000} G \frac{{{m_1}{m_2}}}{{{r^2}}}dx\\ &= G{m_1}{m_2}\left( { - \frac{1}{r}} \right)_R^{R + 1000}\\ &= G{m_1}{m_2}\left( { - \frac{1}{{R + 1000}} + \frac{1}{R}} \right)\end{aligned}\)

That is, \(W = G{m_1}{m_2}\left( {\frac{{1000}}{{R(R + 1000)}}} \right)\)

Substitute the given values to obtain the value of\(W\).

Put the given values and obtain the value of \(W\)as follows.

\(\begin{aligned}{}W &= \left( {6.67 \times {{10}^{ - 11}}} \right)\left( {5.98 \times {{10}^{24}}} \right)(1000)\left( {\frac{{1000}}{{6.37 \times {{10}^6}\left( {6.37 \times {{10}^6} + 1000} \right)}}} \right)\\ &= \frac{{39.8866 \times {{10}^{19}}}}{{6.37 \times {{10}^6}\left( {6.371 \times {{10}^6}} \right)}}n\\ &= 0.983 \times {10^7}\;{\rm{J}}\\ &= 9.83 \times {10^6}\;{\rm{J}}\end{aligned}\)

Therefore, the work needed to boost a satellite of mass\(1000kg\)out of the gravitational field is\(9.83 \times {10^6}J\).