Question

A satellite orbits the Earth at an altitude of 700 km on the equatorial plane of the earth and it revolves in the same direction as the direction of rotation of the earth. Considering the radius of a spherical earth as 6300 km and the acceleration due to gravity as 10 m/s², the tangential velocity of the satellite in the orbit is \(\underline{\hspace{1cm}}\) km/s. (round off to 2 decimal places)

Hint:

The tangential velocity of a satellite in orbit can be calculated using the gravitational constant and the distance from the Earth's center.

Answer 1

Correct Answer: 7.52 - 7.53

The formula for the tangential velocity of a satellite is: \[ v = \sqrt{\frac{GM}{r}}, \] where:
- \( G = 6.674 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2} \) is the gravitational constant,
- \( M = 5.97 \times 10^{24} \, \text{kg} \) is the mass of the Earth,
- \( r \) is the distance from the center of the Earth to the satellite, which is the sum of the Earth's radius and the satellite's altitude:
\[ r = 6300 \, \text{km} + 700 \, \text{km} = 7000 \, \text{km} = 7 \times 10^6 \, \text{m}. \] Now, substituting the values: \[ v = \sqrt{\frac{(6.674 \times 10^{-11}) \times (5.97 \times 10^{24})}{7 \times 10^6}} \approx 7520 \, \text{m/s} = 7.52 \, \text{km/s}. \] Thus, the tangential velocity of the satellite is \( \boxed{7.52} \, \text{km/s} \).