Question

Given satellite mass\[ = 100\,{\text{kg}}\] with circular orbit of radius \[25000\,{\text{km}}\] around the Earth. [Mass of the earth\[ = 6 \times {10^{24}}\,{\text{kg}}\], \[G = 6.67 \times {10^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}{\text{/k}}{{\text{g}}^{\text{2}}}\]]. Find gravitational force between the Earth and satellite.

Answer 1

Hint:Use the expression for Newton’s law of gravitation. This expression gives the relation between the gravitational force of attraction between the two objects, universal gravitational constant, masses of the two objects and distance between the centres of the two objects. Substitute all the given values in this expression and calculate the gravitational force of attraction between the satellite and the Earth.

Formula used:
The expression for Newton’s law of gravitation is given by
\[F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}\] …… (1)
Here, \[F\] is the gravitational force of attraction between the two objects, \[G\] is the universal gravitational constant, \[{m_1}\] and \[{m_2}\] are the masses of the two objects and \[r\] is the distance between the centres of the two objects.

Complete step by step answer:
We have given that the mass of the satellite is \[100\,{\text{kg}}\].
\[m = 100\,{\text{kg}}\]
The radius of the orbit of the satellite around the Earth is \[25000\,{\text{km}}\].
\[r = 25000\,{\text{km}}\]
The mass of the Earth is \[6 \times {10^{24}}\,{\text{kg}}\] and the value of universal gravitational constant is \[6.67 \times {10^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}{\text{/k}}{{\text{g}}^{\text{2}}}\].
\[M = 6 \times {10^{24}}\,{\text{kg}}\]
\[G = 6.67 \times {10^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}{\text{/k}}{{\text{g}}^{\text{2}}}\]
We have asked to calculate the gravitational force of attraction between the satellite and the Earth.Let us first convert the unit of radius of the circular orbit of the satellite around the Earth in the SI system of units.
\[r = \left( {25000\,{\text{km}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{m}}}}{{1\,{\text{km}}}}} \right)\]
\[ \Rightarrow r = 25 \times {10^6}\,{\text{m}}\]
Hence, the radius of the orbit of the satellite around the Earth is \[25 \times {10^6}\,{\text{m}}\].
Rewrite equation (1) for the gravitational force of attraction between the satellite and the Earth.
\[F = \dfrac{{GMm}}{{{r^2}}}\]
Substitute \[6.67 \times {10^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}{\text{/k}}{{\text{g}}^{\text{2}}}\] for \[G\], \[6 \times {10^{24}}\,{\text{kg}}\] for \[M\], \[100\,{\text{kg}}\] for \[m\] and \[25 \times {10^6}\,{\text{m}}\] for \[r\] in the above equation.
\[F = \dfrac{{\left( {6.67 \times {{10}^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}{\text{/k}}{{\text{g}}^{\text{2}}}} \right)\left( {6 \times {{10}^{24}}\,{\text{kg}}} \right)\left( {100\,{\text{kg}}} \right)}}{{{{\left( {25 \times {{10}^6}\,{\text{m}}} \right)}^2}}}\]
\[ \therefore F = 64.032\,{\text{N}}\]

Hence, the gravitational force of attraction between the satellite and the Earth is \[64.032\,{\text{N}}\].

Note:The students should not forget to convert the unit of the radius of the circular orbit of the satellite around the Earth in the SI system of units as units of all the physical quantities in the expression for Newton’s gravitational law are in the SI system of units. If one does not convert this separation distance in the SI system of units then the final answer will be incorrect.