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Chapter 6: Problem 768

A small satellite is revolving near earth's surface. Its orbital velocity will be nearly \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) 8 (B) 4 (C) 6 (D) \(11.2\)

Short Answer

Expert verified
The orbital velocity of a small satellite revolving near the Earth's surface is approximately 8 \(\mathrm{km/s}\).

Step by step solution

01

Write the formula for orbital velocity

Orbital velocity (V) is the velocity an object requires to stay in a stable orbit around another celestial body, like a planet or a moon. The formula for orbital velocity is given by: \[V = \sqrt{\frac{GM}{R}}\] Where: - V is the orbital velocity - G is the gravitational constant (\(6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^2 \cdot \mathrm{kg}^{-2}\)) - M is the mass of the celestial body (in this case, Earth's mass, which is \(5.97 \times 10^{24} \mathrm{kg}\)) - R is the distance from the center of the celestial body to the object in orbit (in this case, Earth's radius, which is approximately \(6.37 \times 10^6 \mathrm{m}\))
02

Substitute values and calculate the orbital velocity

We'll now substitute the given values for G, M, and R into the orbital velocity formula: \[V = \sqrt{\frac{(6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^2 \cdot \mathrm{kg}^{-2})(5.97 \times 10^{24} \mathrm{kg})}{(6.37 \times 10^6 \mathrm{m})}}\] Now we'll solve for V: \[V \approx 7.9 \times 10^3 \mathrm{m/s}\]
03

Convert the orbital velocity to \(\mathrm{km/s}\) and select the correct answer

Wrap up the process by converting the orbital velocity from \(\mathrm{m/s}\) to \(\mathrm{km/s}\) as requested: \[V \approx 7.9 \times 10^3 \mathrm{m/s} \times (\frac{1 \mathrm{km}}{1000 \mathrm{m}}) \approx 7.9 \mathrm{km/s}\] Looking at the given choices, the closest option to our calculated orbital velocity of \(7.9 \mathrm{km/s}\) is choice (A) 8 \(\mathrm{km/s}\). So, the correct answer is: (A) 8 \(\mathrm{km/s}\)

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