Problem 1
Question
The Fig. \(7.9\) shows a planet in elliptical orbit around the sun \(S\). Where is the kinetic energy of the planet maximum? (A) \(P_{1}\) (B) \(P_{2}\) (C) \(P_{3}\) (D) \(P_{4}\)
Step-by-Step Solution
Verified Answer
The kinetic energy of the planet is maximum at the perihelion position, which is the point closest to the sun. Therefore, the correct answer is (A) \(P_{1}\).
1Step 1: Recall Kepler's second law of planetary motion
Kepler's second law states that "A line segment joining a planet and the sun sweeps out equal areas during equal intervals of time". Mathematically, this can be described as:
\(\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt} = \text{constant}\)
Where \(dA\) is the area swept out by the planet in the orbital path, \(dt\) is the time interval, \(r\) is the distance between the planet and the sun, and \(\theta\) is the angle between the planet, the sun, and a fixed point along the orbit.
2Step 2: Derive an expression for kinetic energy
The total mechanical energy of the planet in an orbit is given by:
\(E_\text{total} = E_\text{kinetic} + E_\text{potential}\)
For an elliptical orbit, the gravitational potential energy of the planet is:
\(E_\text{potential} = -\frac{GMm}{r}\)
Where \(G\) is the gravitational constant, \(M\) is the mass of the sun, and \(m\) is the mass of the planet. Since \(\frac{dA}{dt}\) is constant, as \(r\) decreases, the angular velocity \(\frac{d\theta}{dt}\) increases. This implies that the planet's linear velocity increases when its distance from the sun decreases. Therefore, the planet's kinetic energy, \(E_\text{kinetic} = \frac{1}{2} mv^2\), will be maximum when the distance \(r\) is minimum.
3Step 3: Determine the minimum and maximum distances
Given that the orbit is elliptical, the minimum distance between the planet and the sun corresponds to the perihelion position, and the maximum distance corresponds to the aphelion position. In the given figure, \(P_1\) is the perihelion position (closest to the sun), and \(P_3\) is the aphelion position (farthest from the sun).
4Step 4: Identify the point of maximum kinetic energy
Since maximum kinetic energy corresponds to the minimum distance from the sun, the kinetic energy of the planet will be maximum at the perihelion position \(P_1\). Therefore, the correct answer is:
(A) \(P_{1}\)
Other exercises in this chapter
Problem 2
The ratio of the radii of the planets \(P_{1}\) and \(P_{2}\) is \(k_{1}\). The ratio of the acceleration due to the gravity on them is \(k_{2}\). The ratio of
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The orbital speed of Jupiter is (A) greater than the orbital speed of earth. (B) lesser than the orbital speed of earth. (C) equal to the orbital speed of earth
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The period of a satellite in a circular orbit of radius \(R\) is \(T\). The period of another satellite in a circular orbit of radius \(4 R\) is (A) \(4 T\) (B)
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