Question

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 \( M = 6 \times 10^{24} \, \text{kg} \))

• A. \( 7.0 \, \text{km/s} \)
• B. \( 8.0 \, \text{km/s} \)
• C. \( 9.0 \, \text{km/s} \)
• D. \( 10.0 \, \text{km/s} \)

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Use the formula for the orbital speed of a satellite The orbital speed \( v \) of a satellite orbiting at a height \( h \) above the Earth's surface is given by: \[ v = \sqrt{\frac{GM}{r}} \] where: - \( G \) is the gravitational constant, - \( M \) is the mass of the Earth, - \( r \) is the distance from the center of the Earth to the satellite, which is \( r = R + h \), where \( R \) is the radius of the Earth. Step 2: Substitute the given values Given: - \( G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \), - \( M = 6 \times 10^{24} \, \text{kg} \), - Radius of the Earth \( R = 6.4 \times 10^6 \, \text{m} \), - Height of the satellite \( h = 10^4 \, \text{km} = 10^7 \, \text{m} \). The total distance from the center of the Earth to the satellite is: \[ r = 6.4 \times 10^6 + 10^7 = 1.64 \times 10^7 \, \text{m} \] Now, substitute these values 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: Therefore, the orbital speed of the satellite is \( 7.0 \, \text{km/s} \). So, the correct answer is option (1).