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Chapter 6: Problem 81

Prove that \(U=-2 K\) for any gravitational circular orbit. [Hint: Use Newton's second law to relate the gravitational force to the acceleration required to maintain uniform circular motion.]

Short Answer

Expert verified
Question: Prove that the relationship between gravitational potential energy (U) and kinetic energy (K) for any gravitational circular orbit is given by \(U=-2 K\). Solution: To prove that \(U=-2 K\) for any gravitational circular orbit, we need to: 1. Define the expressions for potential energy (U) and kinetic energy (K) in a gravitational system 2. Relate the gravitational force and centripetal force using Newton's second law 3. Substitute the expression for orbital velocity into the kinetic energy formula 4. Show the relationship between potential and kinetic energies by comparing their expressions Following these steps, we find that \(U = - 2K\) for any gravitational circular orbit.

Step by step solution

01

Define potential and kinetic energies

For a gravitational system, the potential energy (U) is given by the formula: \(U = -\frac{GMm}{r}\) Where G is the gravitational constant, M is the mass of the central object (e.x. the Earth), m is the mass of the object in orbit (e.x. a satellite), and r is the distance between their centers of mass. The kinetic energy (K) is given by the formula: \(K = \frac{1}{2}mv^2\) Where v is the orbital velocity of the object in orbit (i.e., the satellite).
02

Relate the gravitational force and centripetal force

According to Newton's second law, in a circular orbit, the gravitational force acts as the centripetal force that keeps the object in orbit. Therefore, we have: \(F_g = F_c\) \(\frac{GMm}{r^2} = m\frac{v^2}{r}\) Now, let's rearrange the formula for v: \(v^2 =\frac{GM}{r}\)
03

Substitute the expression for orbital velocity into the kinetic energy formula

Now, let's substitute the expression for \(v^2\) found in step 2 into the formula for the kinetic energy: \(K = \frac{1}{2}m\frac{GM}{r}\)
04

Show the relationship between potential and kinetic energies

Next, let's compare the expressions for potential energy (U) and kinetic energy (K) to find the relationship between them: \(U = -\frac{GMm}{r}\) \(K = \frac{1}{2}m\frac{GM}{r}\) Notice that if you multiply the equation for K by 2, the right-hand side of the equation will match the right-hand side of the equation for U but with opposite signs: \(2 K = -U\) This proves that for any gravitational circular orbit: \(U = - 2K\)

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