A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ______ x $10^{10}$ J.
(Mass of earth = $6 \times 10^{24}$ kg, Radius of earth = $6.4 \times 10^6$ m, Gravitational constant = $6.67 \times 10^{-11}$ Nm$^2$ kg$^{-2}$)
(Mass of earth = $6 \times 10^{24}$ kg, Radius of earth = $6.4 \times 10^6$ m, Gravitational constant = $6.67 \times 10^{-11}$ Nm$^2$ kg$^{-2}$)
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Correct Answer: 3
Approach Solution - 1
We are asked to find the kinetic energy of a satellite of mass \( m = 1000\,\text{kg} \) revolving around the Earth at a height \( h = 270\,\text{km} \) above the surface.
Given Data:
\[ m = 1000\,\text{kg}, \quad M = 6 \times 10^{24}\,\text{kg}, \quad R = 6.4 \times 10^6\,\text{m}, \quad G = 6.67 \times 10^{-11}\,\text{Nm}^2/\text{kg}^2 \] \[ h = 270\,\text{km} = 270 \times 10^3\,\text{m} \]
Concept Used:
For a satellite in a circular orbit, the gravitational force provides the centripetal force:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \] \[ \Rightarrow v^2 = \frac{GM}{r} \]
where \( r = R + h \) is the orbital radius.
The kinetic energy of the satellite is:
\[ K = \frac{1}{2}mv^2 = \frac{1}{2}m\frac{GM}{r} \]
Step-by-Step Solution:
Step 1: Compute the orbital radius:
\[ r = R + h = 6.4 \times 10^6 + 0.27 \times 10^6 = 6.67 \times 10^6\,\text{m} \]
Step 2: Substitute in the kinetic energy formula:
\[ K = \frac{1}{2}m\frac{GM}{r} \]
Step 3: Substitute numerical values:
\[ K = \frac{1}{2} \times 1000 \times \frac{(6.67 \times 10^{-11})(6 \times 10^{24})}{6.67 \times 10^6} \]
Step 4: Simplify step by step:
\[ \frac{GM}{r} = \frac{6.67 \times 6}{6.67} \times 10^{-11 + 24 - 6} = 6 \times 10^7 \] \[ K = \frac{1}{2} \times 1000 \times 6 \times 10^7 = 3 \times 10^{10}\,\text{J} \]
Final Computation & Result:
The kinetic energy of the satellite is:
\[ \boxed{K = 3 \times 10^{10}\,\text{J}} \]
Hence, the required answer is 3 × 1010 J.
Approach Solution -2
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