Question

A satellite revolves very near to the earth surface. Its speed should be around

• A. 5 km/s
• B. 8 km/s
• C. 2 km/s
• D. 11 km/s

Hint:

First cosmic velocity (orbital) \(\approx 7.9\ \mathrm{km/s}\); second cosmic velocity (escape) \(\approx 11.2\ \mathrm{km/s}\).

Answer 1

The Correct Option is B

The question is about the speed of a satellite revolving very close to the Earth's surface. Let's derive the speed using the concept of circular motion and gravitational force.

For a satellite revolving close to the Earth's surface, the gravitational force provides the necessary centripetal force.

The formula that relates gravitational force to centripetal force is:

\(F_{\text{gravity}} = F_{\text{centripetal}}\)

Where:

• \(F_{\text{gravity}} = \frac{G \cdot M \cdot m}{r^2}\) is the gravitational force.
• \(F_{\text{centripetal}} = \frac{m \cdot v^2}{r}\) is the centripetal force.
• \(G\) is the universal gravitational constant \((6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2)\).
• \(M\) is the mass of the Earth \((5.972 \times 10^{24} \, \text{kg})\).
• \(m\) is the mass of the satellite.
• \(r\) is the radius of Earth \((6.37 \times 10^6 \, \text{m})\).
• \(v\) is the orbital speed of the satellite.

Equating the two forces, we have:

\(\frac{G \cdot M \cdot m}{r^2} = \frac{m \cdot v^2}{r}\)

The mass of the satellite \(m\) cancels out from both sides:

\(\frac{G \cdot M}{r^2} = \frac{v^2}{r}\)

Rearrange to solve for \(v\):

\(v = \sqrt{\frac{G \cdot M}{r}}\)

Substituting the known values:

\(v = \sqrt{\frac{6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \times 5.972 \times 10^{24} \, \text{kg}}{6.37 \times 10^6 \, \text{m}}}\)

Calculating this gives approximately \(7.9 \, \text{km/s}\), which is very close to \(8 \, \text{km/s}\).

Hence, the correct speed for a satellite revolving very near to the Earth's surface is approximately 8 km/s.

Thus, the correct answer is:

8 km/s