Question

Calculate the change in energy of a $500\;{\rm{kg}}$ satellite when it falls from an altitude of $200\;{\rm{km}}$to $199\;{\rm{km}}$. If this change takes place during one orbit, calculate the retarding force on the satellite. (Given: mass of earth = $6 \times {10^{24}}\;{\rm{kg}}$ and radius of earth = $6400\;{\rm{km}}$).

Answer 1

The above problem can be resolved by using the mathematical formula for the change in energy of satellite while moving from any specific point to any height. When satellite orbits around the heavenly objects, such that these keep on changing the vertical height. Thus, with a change in height, the magnitude of energy will also change and will be at its peak, when the satellite is at a higher point.

Complete step by step answer:
Given:
The mass of the satellite is, $m = 500\;{\rm{kg}}$.
The initial height is, ${h_1} = 200\;{\rm{km}}$.
The final height is, ${h_2} = 199\;{\rm{km}}$.
The mass of earth is, $M = 6 \times {10^{24}}\;{\rm{kg}}$.
The radius of earth is, $R = 6400\;{\rm{km}}$.
The mathematical expression for the change in energy of satellite in the given condition is,
$\Delta E = GMm\left( {\dfrac{1}{{R + {h_1}}} - \dfrac{1}{{R + {h_2}}}} \right)$………………………………………….(1)
here, G is the universal gravitational constant and its value is $6.67 \times {10^{ - 11}}\;{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}$.
Solve by substituting the values in equation 1 as,

\Delta E = GMm\left( {\dfrac{1}{{R + {h_1}}} - \dfrac{1}{{R + {h_2}}}} \right)\\\ \Delta E = \left( {6.67 \times {{10}^{ - 11}}\;{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}} \right) \times \left( {6 \times {{10}^{24}}\;{\rm{kg}}} \right) \times \left( {500\;{\rm{kg}}} \right)\left( {\dfrac{1}{{6400\;{\rm{km}} + 200\;{\rm{km}}}} - \dfrac{1}{{6400\;{\rm{km}} + 199\;{\rm{km}}}}} \right)\\\ \Delta E = - 4.59 \times {10^6}\;{\rm{J}} \end{array}$$ The mathematical formula for the retarding force on the satellite is given as, $$F = \dfrac{{\Delta E}}{C}$$……………………………………. (2) Here, C is the circumference of the path of travel of the satellite. And its value is, $$C = 2\pi \left( {R + {h_1}} \right)$$. Solve by substituting the values in equation 2 as, $$\begin{array}{l} F = \dfrac{{\Delta E}}{C}\\\ F = \dfrac{{4.59 \times {{10}^6}\;{\rm{J}}}}{{2\pi \left( {6400\;{\rm{km}} \times \dfrac{{{{10}^3}\;{\rm{m}}}}{{1\;{\rm{km}}}} + 200\;{\rm{km}} \times \dfrac{{{{10}^3}\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)}}\\\ F = 0.11\;{\rm{N}} \end{array}$$ Therefore, the retarding force acting on the satellite is $$0.11\;{\rm{N}}$$. **Note:** To solve the given problem, one must remember the mathematical formula along with the concept of energy change with height for a satellite. Then it is also important to remember the basic formula for the retarding force, which is numerically equal to the fraction of change in energy of the satellite and the circumference of the path of orbit.