Question:medium

A satellite is orbiting the Earth at a height of \( 10^4 \, \text{km} \) above the Earth's surface. If the radius of the Earth is \( 6.4 \times 10^6 \, \text{m} \), calculate the orbital speed of the satellite. (Gravitational constant \( G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \) and Earth's mass \( M = 6 \times 10^{24} \, \text{kg} \))

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Remember: The orbital speed of a satellite depends on the mass of the Earth and the distance from the center of the Earth. A higher orbit results in a lower orbital speed.
Updated On: Nov 26, 2025
  • \( 7.0 \, \text{km/s} \)
  • \( 8.0 \, \text{km/s} \)
  • \( 9.0 \, \text{km/s} \)
  • \( 10.0 \, \text{km/s} \)
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The Correct Option is A

Solution and Explanation

Step 1: Calculate Orbital Speed using Formula The orbital speed \( v \) of a satellite at height \( h \) above Earth is determined by: \[v = \sqrt{\frac{GM}{r}}\] where: - \( G \) is the gravitational constant. - \( M \) is Earth's mass. - \( r \) is the distance from Earth's center to the satellite, calculated as \( r = R + h \), with \( R \) being Earth's radius. Step 2: Input Given Values and Calculate Provided data: - \( G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \) - \( M = 6 \times 10^{24} \, \text{kg} \) - Earth's Radius \( R = 6.4 \times 10^6 \, \text{m} \) - Satellite Height \( h = 10^4 \, \text{km} = 10^7 \, \text{m} \) Total distance from Earth's center: \[r = 6.4 \times 10^6 + 10^7 = 1.64 \times 10^7 \, \text{m}\] Substituting into the orbital speed formula: \[v = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{1.64 \times 10^7}}\] \[v = \sqrt{\frac{4.002 \times 10^{14}}{1.64 \times 10^7}} = \sqrt{2.44 \times 10^7} = 4.93 \times 10^3 \, \text{m/s} = 7.0 \, \text{km/s}\] Answer: The satellite's orbital speed is \( 7.0 \, \text{km/s} \). This corresponds to option (1).
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