Question

(a) Find the work done to move the body away from a fixed position with the distance\(r = a\) to \(r = b\)

(b) Find the work done to launch a \(1000kg\)satellite vertically to a height of\(1000km\).

Short answer

(a) The work done to move a body away from a fixed position with the distance \(r = a\)to \(r = b\)is\(G{m_1}{m_2}\left( {\frac{1}{a} - \frac{1}{b}} \right)\).

(b) The work done to launch a \(1000\;{\rm{kg }}\)satellite vertically to a height of\(1000\;{\rm{km}}\)\(8.5 \times {10^9}\;{\rm{J}}{\rm{.}}\)

Step-by-step solution

Newton’s law of gravitation.

(a)

Apply Newton’s law of gravitation:

The two bodies with masses\(m1\)and\(m2\)attract each other with a force. It is expressed as follows:

\(F = G\frac{{{m_1}{m_2}}}{{{r^2}}}\) …… (1)

Where\(r\)is the distance between the bodies and \(G\)is the gravitational constant.

Use the Newton’s law of gravitation for calculation.

Write the expression to find the work done\(W\)to move a body from\(a\)to\(b\).

\(\begin{aligned}W &= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n f \left( {x_i^*} \right)\Delta x\\ &= \int_a^b f (x)dx\end{aligned}\) …… (2)

Consider the mass of two bodies are\(m1\)and\(m2\).

Consider the mass\(m1\)is fixed and mass\(m2\)is moving.

According to Newton's Law of Gravitation, the two masses \(m1\)and \(m2\)attract each other with a force\(F\).

Calculate the work done to move the mass \(m2\)from a location \(r = a\)to \(r = b\) (against the attraction force of mass\(m1\)) using Equation (2).

Substitute \(F\)for \(f(x)\)and \(dr\)for \(dx\)in Equation (2).

\(W = \int_a^b F dr\) …… (3)

Modify equation (3) using equation (1).

\(W = \int_a^b {\left( {G\frac{{{m_1}{m_2}}}{{{r^2}}}} \right)} dr\)

\( = G{m_1}{m_2}\int_a^b {{r^{ - 2}}} dr\) ...... (4)

\( = G{m_1}{m_2}\left[ { - \frac{1}{r}} \right]_a^b\)

\( = G{m_1}{m_2}\left( {\frac{1}{a} - \frac{1}{b}} \right)\)

Thus, the work done to move the mass\(m2\)from the location\(r = a\) to\(r = b\), against the attraction force of mass\(m1\)is\(G{m_1}{m_2}\left( {\frac{1}{a} - \frac{1}{b}} \right)\).

Assume the mass of the earth \({m_1}{\rm{ as }}5.98 \times {10^{24}}\;{\rm{kg}}\)for calculation.

(b)

Assume the mass of the earth \({m_1}{\rm{ as }}5.98 \times {10^{24}}\;{\rm{kg}}\) and it is concentrated at its centre.

Take the radius of earth\(r{\rm{ as }}6.37 \times {10^6}\;{\rm{m}}\).

Take the value of gravitational constant\(G{\rm{ as }}6.67 \times {10^{ - 11}}\;{\rm{N}} - {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}\).

Take the mass of the satellite\({m_2}{\rm{ as }}1000\;{\rm{kg}}\).

The mass of the earth is concentrated at its centre.

The satellite is to move from the surface of the earth to a height of\(1000km.\)

Calculate the value of\(a\) as shown below.

\(a = \)radius of the earth

\( = 6.37 \times {10^6}\;{\rm{m}}\)

Take the value of\(b\)as shown below.

\(b = \)radius of the earth + height of lift of satellite.

\(\begin{aligned}{} &= 6.37 \times {10^6}\;{\rm{m}} + 1000\;{\rm{km}} \times \left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\\ &= \left( {6.37 \times {{10}^6} + 1 \times {{10}^6}\;{\rm{m}}} \right)\\ &= 7.37 \times {10^6}\;{\rm{m}}\end{aligned}\)

Calculate the work done\(W\).

Calculate the work done\(W\)to lift the satellite to a height of\(1000km\), using equation (4) as shown below:

Substitute

\(6.67 \times {10^{ - 11}}\;{{\rm{N}}_{ - {{\rm{m}}^2}}}/{\rm{k}}{{\rm{g}}^2}{\rm{ for G}},5.98 \times {10^{24}}\;{\rm{kg for }}{m_1},1000\;{\rm{kg for }}{m_2},6.37 \times {10^6}\;{\rm{m for }}a\), and \(7.37 \times {10^6}\;{\rm{m for }}b\)in equation(4).

\(\begin{aligned}W &= G{m_1}{m_2}\left( {\frac{1}{a} - \frac{1}{b}} \right)\\ &= 6.67 \times {10^{ - 11}} \times 5.98 \times {10^{24}} \times 1000\left( {\frac{1}{{6.37 \times {{10}^6}}} - \frac{1}{{7.37 \times {{10}^6}}}} \right)\\ &= 6.67 \times 5.98 \times 2.13 \times {10^9}\\ &= 8.5 \times {10^9}\;{\rm{J}}\end{aligned}\)

Thus, the work done to launch a \(1000\;{\rm{kg }}\)satellite vertically to a height of\(1000\;{\rm{km}}\)\(8.5 \times {10^9}\;{\rm{J}}{\rm{.}}\)