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Chapter 6: Problem 815

An artificial satellite moving in a circular orbit around earth has a total (kinetic + potential energy) \(E_{0}\), its potential energy is (A) \(-\mathrm{E}_{0}\) (B) \(1.5 \mathrm{E}_{0}\) (C) \(2 \mathrm{E}_{0}\) (D) \(\mathrm{E}_{0}\)

Short Answer

Expert verified
The potential energy (U) of an artificial satellite moving in a circular orbit around Earth is twice the total energy (E): \(U = -2E_{0}\). The correct option is (C) \(2 \mathrm{E}_{0}\).

Step by step solution

01

Recall the formulas for gravitational potential energy and kinetic energy

Gravitational potential energy (U) is given by the formula: \(U = -\dfrac{GMm}{r}\), where G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance between the centers of the Earth and the satellite. Kinetic energy (K) is given by the formula: \(K = \dfrac{1}{2}mv^2\), where m is the mass of the satellite, and v is its orbital velocity.
02

Express the orbital velocity in terms of the gravitational constant, Earth's mass, and distance

For a satellite in a circular orbit, the centripetal force is balanced by the gravitational force. That is, \(m\dfrac{v^2}{r} = \dfrac{GMm}{r^2}\). We can rearrange this equation for the orbital velocity: \(v^2 = \dfrac{GM}{r}\).
03

Substitute the orbital velocity into the kinetic energy formula

Now, use the expression for orbital velocity found in Step 2 to substitute into the kinetic energy formula: \(K = \dfrac{1}{2}m\left(\dfrac{GM}{r}\right)\).
04

Find the total energy (kinetic and potential energy) of the satellite

Add the kinetic and potential energy formulas to find the total energy of the satellite (E): \(E_{0} = U + K = -\dfrac{GMm}{r} + \dfrac{1}{2}m\left(\dfrac{GM}{r}\right)\).
05

Solve for potential energy in terms of total energy

We have \(E_{0} = -\dfrac{GMm}{r} + \dfrac{1}{2}m\left(\dfrac{GM}{r}\right)\). Simplifying this equation, we get: \(E_{0} = - \dfrac{1}{2}\dfrac{GMm}{r}\). Thus, potential energy (U) in terms of total energy (E) is twice of the total energy: \(U = -2E_{0}\).
06

Choose the correct answer from the given options

From the result obtained in Step 5, the potential energy (U) is -2 times the total energy (E). Hence, the correct option is (C) \(2 \mathrm{E}_{0}\).

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Most popular questions from this chapter

If the radius of the earth were to shrink by \(1 \%\) its mass remaining the same, the acceleration due to gravity on the earth's surface would (A) decrease by \(2 \%\) (B) remain Unchanged (C) increase by \(2 \%\) (D) increases by \(1 \%\)

The escape velocity for a body projected vertically upwards from the surface of earth is \(11 \mathrm{kms}^{-1}\). If the body is projected at an angle of \(45^{\circ}\) with the vertical, the escape velocity will be $\ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \((11 / \sqrt{2})\) (B) \(11 \sqrt{2}\) (C) 22 (D) 11

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The additional K.E. to be provided to a satellite of mass \(\mathrm{m}\) revolving around a planet of mass \(\mathrm{M}\), to transfer it from a circular orbit of radius \(\mathrm{R}_{1}\) to another radius \(\mathrm{R}_{2}\left(\mathrm{R}_{2}>\mathrm{R}_{1}\right)\) is (A) $\operatorname{GMm}\left[\left(1 / R_{1}^{2}\right)-\left(1 / R_{2}^{2}\right)\right]$ \(\operatorname{GMm}\left[\left(1 / R_{1}\right)-\left(1 / R_{2}\right)\right]\) (C) $2 \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$ (D) $(1 / 2) \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$

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