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Chapter 6: Problem 729

Radius of orbit of satellite of earth is \(\mathrm{R}\). Its \(\mathrm{KE}\) is proportional to (A) \((1 / R)\) (B) \((1 / \sqrt{\mathrm{R}})\) (C) \(\mathrm{R}\) (D) \(\left(1 / \mathrm{R}^{3 / 2}\right)\)

Short Answer

Expert verified
The kinetic energy (KE) of a satellite orbiting Earth is given by the expression \(KE = \frac{1}{2} m_s * v^2\), where \(m_s\) is the mass of the satellite and \(v\) is the orbital speed. Using the relationship between gravitational force and centripetal force, we find that \(v^2 = \frac{G * m_e}{R}\), where \(G\) is the gravitational constant, \(m_e\) is the mass of Earth, and \(R\) is the radius of the orbit. Substituting this expression into the kinetic energy formula, we get \(KE = \frac{1}{2} m_s * \frac{G * m_e}{R}\), which simplifies to \(KE \propto \frac{1}{R}\). Therefore, the correct answer is (A) \(\frac{1}{R}\).

Step by step solution

01

Write the expression for gravitational force

The gravitational force between Earth and the satellite can be expressed as: \[F_G = \frac{G * m_e * m_s}{R^2}\] where \(F_G\) is the gravitational force, \(G\) is the gravitational constant, \(m_e\) is the mass of Earth, and \(m_s\) is the mass of the satellite.
02

Write the expression for the centripetal force

The centripetal force required for the satellite's circular motion can be written as: \[F_c = \frac{m_s * v^2}{R}\] where \(F_c\) is the centripetal force and \(v\) is the satellite's orbital speed.
03

Equate gravitational force and centripetal force

Since the gravitational force between Earth and the satellite provides the centripetal force required for the satellite's circular motion, we can set the two expressions equal: \[\frac{G * m_e * m_s}{R^2} = \frac{m_s * v^2}{R}\]
04

Solve for the satellite's orbital speed

To find the satellite's orbital speed, solve the equation above for \(v^2\): \[v^2 = \frac{G * m_e}{R}\]
05

Write the expression for kinetic energy

The kinetic energy (KE) of the satellite orbiting Earth is given by the expression: \[KE = \frac{1}{2} m_s * v^2\]
06

Substitute the expression for orbital speed into the kinetic energy expression

By substituting the expression from step 4 into the kinetic energy expression, we get: \[KE = \frac{1}{2} m_s * \frac{G * m_e}{R}\]
07

Determine how kinetic energy is proportional to \(R\)

Simplifying the expression from step 6, we have: \[KE \propto \frac{1}{R}\] which means that the kinetic energy of the satellite is proportional to \(\frac{1}{R}\).
08

Select the correct answer

Based on the above derivation, the correct answer is: (A) \(\frac{1}{R}\)

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