Question

A planet with a mass of \(7.00 \cdot 10^{21} \mathrm{~kg}\) is in a circular orbit around a star with a mass of \(2.00 \cdot 10^{30} \mathrm{~kg} .\) The planet has an orbital radius of \(3.00 \cdot 10^{10} \mathrm{~m}\). a) What is the linear orbital velocity of the planet? b) What is the period of the planet's orbit? c) What is the total mechanical energy of the planet?

Short answer

Question: A planet with mass m_p orbits around a star with mass m_s. The distance between the planet and the star is r. Find the linear orbital velocity (v), the period of the planet's orbit (T), and the total mechanical energy (E) of the planet. Answer: To find the linear orbital velocity (v), the period of the planet's orbit (T), and the total mechanical energy (E) of the planet, follow these steps: 1. Find the gravitational force between the planet and the star using Newton's law of universal gravitation: \(F = G\frac{m_pm_s}{r^2}\) 2. Determine the centripetal force keeping the planet in a circular orbit: \(F_c = m_pv^2/r\) 3. Set the equations from steps 1 and 2 equal to each other and solve for v: \(v=\sqrt{\frac{Gm_s}{r}}\) 4. Calculate the period of the planet's orbit (T): \(T=\frac{2\pi r}{v}\) 5. Calculate the total mechanical energy of the planet (E): \(E=KE + PE = \frac{1}{2}m_pv^2 - G\frac{m_pm_s}{r}\) Using the given values for m_p, m_s, and r, along with the gravitational constant G, calculate the linear orbital velocity, the period of the planet's orbit, and the total mechanical energy of the planet.

Step-by-step solution

Gravitational force

First, we want to find the gravitational force between the planet and the star. This can be done using Newton's law of universal gravitation: \(F = G\frac{m_pm_s}{r^2}\) Where \(F\) is the gravitational force, \(G\) is the gravitational constant (\(6.67 \cdot 10^{-11} \mathrm{N(m/kg)^2}\)), \(m_p\) is the mass of the planet, \(m_s\) is the mass of the star, and \(r\) is the distance between them.

Centripetal force

The centripetal force keeping the planet in a circular orbit is equal in magnitude to the gravitational force. The centripetal force can be expressed as: \(F_c = m_pv^2/r\) Where \(F_c\) is the centripetal force, \(m_p\) is the mass of the planet, \(v\) is the linear velocity, and \(r\) is the orbital radius.

Solve for linear orbital velocity

Since the gravitational force and centripetal force are equal, we can set the equations from steps 1 and 2 equal to each other and solve for \(v\): \(G\frac{m_pm_s}{r^2}=m_pv^2/r \Rightarrow v=\sqrt{\frac{Gm_s}{r}}\) Now, plug in the given values for \(G\), \(m_s\), and \(r\) to find the linear orbital velocity, \(v\).

Calculate the period of the planet's orbit

To find the period of the planet's orbit, we need to find the time it takes for the planet to travel around the circumference of its orbit, \(2\pi r\). We can find the period, \(T\), by dividing this distance by the linear orbital velocity, \(v\): \(T=\frac{2\pi r}{v}\) Using the linear orbital velocity found in step 3, calculate the period of the planet's orbit.

Calculate total mechanical energy

The total mechanical energy of the planet is the sum of its kinetic energy and gravitational potential energy. The kinetic energy, \(KE\), can be calculated as: \(KE=\frac{1}{2}m_pv^2\) The gravitational potential energy, \(PE\), can be calculated as: \(PE=-G\frac{m_pm_s}{r}\) The total mechanical energy, \(E\), is the sum of the kinetic energy and potential energy: \(E=KE + PE\) Calculate the kinetic energy and potential energy using the values found in step 3 and the given masses. Then, add the kinetic and potential energies together to find the total mechanical energy.

Key concepts

Gravitational Force

In the realm of orbital mechanics, the concept of gravitational force is fundamental. In our case, we look at the gravitational force as the attraction between a planet and a star. This force is calculated using Newton’s law of universal gravitation, which states: \[F = G \frac{m_p m_s}{r^2}\]where:

• \(F\) is the gravitational force,


• \(G\) is the gravitational constant \(6.67 \cdot 10^{-11} \mathrm{N(m/kg)^2}\),


• \(m_p\) is the mass of the planet,


• \(m_s\) is the mass of the star, and


• \(r\) is the distance between the two celestial bodies.

This force is what holds the planet in orbit. The greater the masses or the closer they are, the stronger the gravitational pull. This force provides the necessary centripetal force to maintain the circular orbital path of the planet.

Centripetal Force

To keep a planet in a circular orbit, a centripetal force is required. This is the force that keeps the planet moving in a circle rather than flying off into space. In this context, the centripetal force is provided by the gravitational pull between the planet and the star. We can express it as:\[F_c = m_p \frac{v^2}{r}\]where:

• \(F_c\) is the centripetal force,


• \(v\) is the orbital velocity,


• \(r\) is the radius of the orbit, and


• \(m_p\) is the mass of the planet.

The crucial point here is that the gravitational force is equal in magnitude to the centripetal force required for circular motion. By setting \[G\frac{m_pm_s}{r^2} = m_p\frac{v^2}{r}\]we can relate these quantities, allowing us to solve for unknowns like orbital speed (\(v\)) and orbital period.

Kinetic Energy

Kinetic energy is the energy of motion. For a planet revolving around a star, this energy is a function of its speed in orbit. The expression for kinetic energy is:\[KE = \frac{1}{2} m_p v^2\]where:

• \(KE\) is kinetic energy,


• \(m_p\) is the mass of the planet, and


• \(v\) is the linear velocity of the planet.

In a stable orbit, the planet's kinetic energy arises from the gravitational force acting as a centripetal force. Higher velocities mean more kinetic energy. This energy is balanced against the planet’s gravitational potential energy to maintain a stable orbit.

Potential Energy

In orbital mechanics, gravitational potential energy is significant. It is the energy stored by the virtue of the position of the planet. The formula for potential energy in the context of a planet orbiting a star is given by:\[PE = -G \frac{m_p m_s}{r}\]The negative sign indicates that the potential energy is lower when the planet is closer to the star, reflecting the attractive nature of gravity. Here:

• \(PE\) is the potential energy,


• \(G\) is the gravitational constant,


• \(m_p\) the mass of the planet,


• \(m_s\) is the mass of the star, and


• \(r\) is the orbital radius.

This form of energy counteracts kinetic energy to influence the planet's motion. The total mechanical energy of the planet is the sum of its kinetic and potential energies, which remains conserved unless external forces work on the system.