Multiple Choice
Pluto travels in a fairly elliptical orbit around the Sun. At its closest distance of km, its orbital speed is 6.12 km/s. At its farthest, its orbital speed reduces to just 3.71 km/s. How far is it from the Sun at this point (in km)?
8. Centripetal Forces & Gravitation
Energy of Elliptical Orbits
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A comet is in a highly elliptical orbit around the Sun with a period of 2400 years. If this comet has a mass of approximately kg, what is the total energy of its orbit?
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Verified step by step guidance1
Start by understanding that the total energy of an orbiting body in a gravitational field is the sum of its kinetic and potential energy. For an elliptical orbit, the total energy is constant and can be expressed as: \( E = -\frac{GMm}{2a} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Sun, \( m \) is the mass of the comet, and \( a \) is the semi-major axis of the orbit.
Use Kepler's Third Law to relate the period of the orbit to the semi-major axis. Kepler's Third Law states: \( T^2 = \frac{4\pi^2}{GM}a^3 \), where \( T \) is the period of the orbit. Rearrange this formula to solve for \( a \): \( a = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3} \).
Substitute the given values into the rearranged Kepler's Third Law formula to find the semi-major axis \( a \). Use \( T = 2400 \) years, converting it into seconds, and \( G = 6.674 \times 10^{-11} \text{ m}^3\text{kg}^{-1}\text{s}^{-2} \), \( M = 1.989 \times 10^{30} \text{ kg} \).
Once you have calculated the semi-major axis \( a \), substitute \( a \), \( G \), \( M \), and \( m = 5 \times 10^{14} \text{ kg} \) into the total energy formula \( E = -\frac{GMm}{2a} \) to find the total energy of the comet's orbit.
Interpret the sign of the total energy. A negative total energy indicates a bound orbit, which is typical for elliptical orbits. The magnitude of the energy gives insight into the stability and characteristics of the orbit.
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