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

Answer: The period of the planet's orbit is approximately \(6.47 \cdot 10^6\) seconds. The total mechanical energy of the planet is approximately \(-1.54 \cdot 10^{33}\) joules.

Step-by-step solution

Find the Gravitational Force Between the Planet and Star

To determine the gravitational force between the planet and star, we can use Newton's Law of Universal Gravitation: \(F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}\) Where: \(F_G\) = gravitational force \(G\) = gravitational constant, approximately \(6.674 \cdot 10^{-11} N \cdot m^2 / kg^2\) \(m_1\) = mass of the planet, \(7.00 \cdot 10^{21} kg\) \(m_2\) = mass of the star, \(2.00 \cdot 10^{30} kg\) \(r\) = distance between the centers of the planet and star (orbital radius), \(3.00 \cdot 10^{10} m\)

Calculate the Centripetal Force on the Planet

The centripetal force acting on the planet is equal to the gravitational force we calculated in Step 1: \(F_c = F_G\) Where: \(F_c\) = centripetal force Since centripetal force is also given by the formula: \(F_c = m_1 \cdot \frac{v^2}{r}\) Where: \(v\) = linear orbital velocity We can now set up the equation to solve for the linear orbital velocity of the planet.

Determine the Linear Orbital Velocity of the Planet

Equating the centripetal force and gravitational force, and solving for the linear orbital velocity, we get: \(m_1 \cdot \frac{v^2}{r} = G \cdot \frac{m_1 \cdot m_2}{r^2}\) \(v^2 = \frac{G \cdot m_2}{r}\) \(v = \sqrt{\frac{G \cdot m_2}{r}}\) Plugging in the values, we get: \(v = \sqrt{\frac{(6.674 \cdot 10^{-11})(2.00 \cdot 10^{30})}{(3.00 \cdot 10^{10})}}\) \(v \approx 2.90 \cdot 10^4 m/s\) The linear orbital velocity of the planet is approximately \(2.90 \cdot 10^4 m/s\).

Calculate the Period of the Planet's Orbit

The period of the planet's orbit is the time taken to complete one orbit. We can use the formula: \(T = \frac{2 \pi r}{v}\) Where: \(T\) = period of the orbit Plugging in the values, we get: \(T = \frac{2 \pi (3.00 \cdot 10^{10})}{2.90 \cdot 10^4}\) \(T \approx 6.47 \cdot 10^6 seconds\) The period of the planet's orbit is approximately \(6.47 \cdot 10^6\) seconds.

Calculate the Total Mechanical Energy of the Planet

The total mechanical energy of the planet consists of its potential and kinetic energies. The potential energy can be calculated using the gravitational formula: \(U = -G \cdot \frac{m_1 \cdot m_2}{r}\) The kinetic energy can be calculated using the following formula: \(K = \frac{1}{2} \cdot m_1 \cdot v^2\) The total mechanical energy is the sum of potential and kinetic energies: \(E = U + K\) Plugging in the values, we get: \(E = -G \cdot \frac{(7.00 \cdot 10^{21})(2.00 \cdot 10^{30})}{(3.00 \cdot 10^{10})} + \frac{1}{2} \cdot (7.00 \cdot 10^{21}) \cdot (2.90 \cdot 10^4)^2\) \(E \approx -1.54 \cdot 10^{33} J\) The total mechanical energy of the planet is approximately \(-1.54 \cdot 10^{33}\) joules.