Question:
A satellite is revolving in a circular orbit around the earth with a speed of 8.4 km s\(^{-1}\) at a height where the acceleration due to gravity is 8.4 m s\(^{-2}\). The height of the satellite from the Earth's surface is (Radius of Earth = 6400 km)
A satellite is revolving in a circular orbit around the earth with a speed of 8.4 km s\(^{-1}\) at a height where the acceleration due to gravity is 8.4 m s\(^{-2}\). The height of the satellite from the Earth's surface is (Radius of Earth = 6400 km)
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In satellite motion, always relate \(v^2 = g'r\) and subtract Earth radius to get height.
Updated On: Jun 20, 2026
- 1000 km
- 1400 km
- 2000 km
- 2400 km
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The Correct Option is C
Solution and Explanation
Step 1: Use orbital velocity relation.
For circular orbit: \[ v^2 = g' r \] where \(g' = 8.4 \, \text{m/s}^2\).
Step 2: Convert velocity.
\[ v = 8.4 \, \text{km/s} = 8400 \, \text{m/s} \]
Step 3: Find orbital radius.
\[ r = \frac{v^2}{g'} = \frac{(8400)^2}{8.4} \] \[ = \frac{70560000}{8.4} = 8400000 \, \text{m} \] \[ r = 8400 \, \text{km} \]
Step 4: Find height above Earth.
\[ h = r - R \] \[ h = 8400 - 6400 = 2000 \, \text{km} \]
Step 5: Final conclusion.
\[ \boxed{2000 \, \text{km}} \]
For circular orbit: \[ v^2 = g' r \] where \(g' = 8.4 \, \text{m/s}^2\).
Step 2: Convert velocity.
\[ v = 8.4 \, \text{km/s} = 8400 \, \text{m/s} \]
Step 3: Find orbital radius.
\[ r = \frac{v^2}{g'} = \frac{(8400)^2}{8.4} \] \[ = \frac{70560000}{8.4} = 8400000 \, \text{m} \] \[ r = 8400 \, \text{km} \]
Step 4: Find height above Earth.
\[ h = r - R \] \[ h = 8400 - 6400 = 2000 \, \text{km} \]
Step 5: Final conclusion.
\[ \boxed{2000 \, \text{km}} \]
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