Problem 72 - University Physics with Modern Physics

Question

The motion of a planet in a circular orbit about a star obeys the equations of simple harmonic motion. If the orbit is observed edge-on, so the planet's motion appears to be onedimensional, the analogy is quite direct: The motion of the planet looks just like the motion of an object on a spring a) Use Kepler's Third Law of planetary motion to determine the "spring constant" for a planet in circular orbit around a star with period \(T\) b) When the planet is at the extremes of its motion observed edge-on, the analogous "spring" is extended to its largest displacement. Using the "spring" analogy, determine the orbital velocity of the planet.

Step-by-Step Solution

Answer

Answer: The "spring constant" for a planet in a circular orbit can be expressed as \(kx = M_p\frac{G M_s}{r^2}\), where \(k\) is the spring constant, \(x\) is the displacement, \(M_p\) is the mass of the planet, \(G\) is the gravitational constant, \(M_s\) is the mass of the star, and \(r\) is the radius of the orbit. When the planet is observed edge-on at the extremes of its motion, its orbital velocity remains unchanged and is given by the expression \(v^2 = G\frac{M_s}{r}\).

Step 1: Kepler's Third Law and deriving "spring constant"

According to Kepler's Third Law, the square of the period of a planet is proportional to the cube of the semi-major axis of its orbit. Mathematically, this can be written as: $$ T^2 \propto a^3 $$ where \(T\) is the period of the orbit, and \(a\) is the semi-major axis of the orbit. For simplicity, let's consider the orbit to be circular, so \(a\) becomes the radius \(r\). To find the proportionality constant, we can use the gravitational force formula: $$ F_g = G\frac{M_p M_s}{r^2} $$ where \(F_g\) is the gravitational force between the planet (\(M_p\)) and the star (\(M_s\)), and \(G\) is the gravitational constant. Since the planet is following a circular orbit, the centripetal force is equal to the gravitational force. The centripetal force formula is given by: $$ F_c = M_p\frac{v^2}{r} $$ By equating both forces, we can find the orbital velocity \(v\): $$ G\frac{M_p M_s}{r^2} = M_p\frac{v^2}{r} $$ Now, let's find an expression for the "spring constant" using the analogy with Hooke's Law. We know that the force exerted by a spring is given by: $$ F = -kx $$ where \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position. Equating this to the centripetal force, we get: $$ kx = M_p\frac{v^2}{r} $$ Now, we want to express \(k\) in terms of the orbital period \(T\). To do this, we'll first find an expression for the orbital velocity \(v\) in terms of \(T\) from the equation we derived earlier: $$ G\frac{M_p M_s}{r^2} = M_p\frac{v^2}{r} $$ From this equation, we can find that: $$ v^2 = G\frac{M_s}{r} $$ By substituting this expression for velocity into the equation for the "spring constant," we can find \(k\) in terms of \(T\): $$ kx = M_p\frac{G M_s}{r^2} $$

Step 2: Finding the orbital velocity at the extremes of motion

From the "spring constant" equation: $$ kx = M_p\frac{G M_s}{r^2} $$ At the extremes of motion, the displacement \(x\) is equal to the radius \(r\). Hence, we can substitute \(x\) by \(r\): $$ kr = M_p\frac{G M_s}{r^2} $$ Now, we can solve for the orbital velocity \(v\) at this extreme point by recalling the centripetal force equation: $$ M_p\frac{v^2}{r} = kr $$ Substituting the "spring constant" we derived before: $$ M_p\frac{v^2}{r} = M_p\frac{G M_s}{r^2} $$ Finally, we find the orbital velocity when the planet is at the extremes of its motion: $$ v^2 = G\frac{M_s}{r} $$ which is the same expression we derived for the orbital velocity in general. Therefore, the orbital velocity of the planet at the extremes of its motion observed edge-on remains unchanged. In conclusion, the "spring constant" for a planet in a circular orbit with period \(T\) can be determined using Kepler's Third Law, and the orbital velocity of the planet remains the same at the extremes of its motion observed edge-on.

Key Concepts

Understanding Simple Harmonic Motion

Simple harmonic motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. Objects that exhibit simple harmonic motion oscillate about an equilibrium position, the location where the object naturally comes to rest.

One of the classic examples of SHM is a mass attached to a spring, which, when displaced from its rest position, experiences a restoring force courtesy of Hooke's Law. This law states that the force exerted by the spring is proportional to the displacement:
$$ F = -kx $$where: \(F\) is the force applied, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position.

The negative sign indicates that the force is in the opposite direction of the displacement. In the context of our planet's motion, when looked at edge-on, it behaves analogously to an object in simple harmonic motion, where the gravitational pull provides the restoring force.

Gravitational Force Formula and Kepler's Laws

According to Newton's law of universal gravitation, the force between two masses is given by the formula:$$ F_g = G\frac{M_pM_s}{r^2} $$where \(F_g\) is the gravitational force, \(G\) is the universal gravitational constant, \(M_p\) and \(M_s\) are the masses of the planet and the star respectively, and \(r\) is the distance between their centers.

Kepler's Third Law provides a relationship between the orbiting period of a planet and its distance from the star. It states that the square of the orbital period (\(T\)) of a planet is proportional to the cube of the semi-major axis of its orbit (which is the radius \(r\) in the case of a circular orbit): $$ T^2 \propto r^3 $$ By understanding this relationship, we can infer attributes of the orbit and the forces acting upon a planet, aiding us in calculating the equivalent 'spring constant' for the planet's orbital motion.

Centripetal Force: The Force behind Circular Motion

Centripetal force is the required force that keeps an object moving in a circular path and is directed towards the center around which the object is moving. It can be described by the equation:$$ F_c = M_p\frac{v^2}{r} $$Here, \(F_c\) denotes the centripetal force, \(M_p\) is the mass of the orbiting object, \(v\) is the velocity of the object, and \(r\) is the radius of the circular orbit.

In the scenario of a planet orbiting a star, the gravitational force acts as the centripetal force that keeps the planet in its circular path. This is why the planet's orbit can be analyzed using principles of both simple harmonic motion and circular motion. The planet's speed at the furthest points in its orbit can be determined through the balance of gravitational and centripetal forces, showing that despite the planet's variation in position, its velocity remains constant when observed edge-on.