Problem 812
Question
Suppose the gravitational force varies inversely as the nth power of distance the time period of planet in circular orbit of radius \(\mathrm{R}\) around the sun will be proportional to (A) \(\mathrm{R}^{[(\mathrm{n}+1) / 2]}\) (B) \(\mathrm{R}^{[(\mathrm{n}-1) / 2]}\) (C) \(\mathrm{R}^{\mathrm{n}}\) (D) \(\mathrm{R}^{[(\mathrm{n}-1) / 2]}\)
Step-by-Step Solution
Verified Answer
The time period (T) of a planet in a circular orbit of radius R around the sun, when the gravitational force varies inversely as the nth power of distance, is proportional to \(R^{[(n+1) / 2]}\). Therefore, the correct answer is (A) \(R^{[(n+1) / 2]}\).
1Step 1: Recall Kepler's Third Law
Kepler's third law states that the square of the time period of a planet's orbit around the sun is proportional to the cube of the semi-major axis of the orbit. For a circular orbit, the semi-major axis is equal to the radius of the orbit. Mathematically, this can be written as:
\(T^2 = k R^3\),
where T is the time period, R is the radius of the orbit, and k is the constant of proportionality.
2Step 2: Modify Kepler's Third Law for the given force variation
According to the problem, the gravitational force varies inversely as the nth power of distance, which can be written as:
\(F = \frac{G M m}{R^n}\),
where F is the gravitational force, G is the gravitational constant, M is the mass of the sun, m is the mass of the planet, and n is the power of distance dependence.
3Step 3: Express centripetal force in terms of radius
Centripetal force is required for the planet to stay in a circular orbit around the sun. This force is equal to the gravitational force. Therefore, we can write:
\(\frac{G M m}{R^n} = \frac{m v^2}{R}\),
where v is the tangential velocity of the planet.
Now, we can rearrange this equation to find the relationship between v and R:
\(v^2 = \frac{G M R^{(n-2)}}{R^n} = G M R^{(2-n)}\).
4Step 4: Relate time period, velocity and radius
The time period can also be defined as the time taken for the planet to complete one full orbit around the sun, which can be defined as:
\(T = \frac{2 \pi R}{v}\),
Now, square T to find the relation between T and R:
\(T^2 = (\frac{2 \pi R}{v})^2\).
5Step 5: Derive the relationship between time period and radius
Now, substitute the expression for v^2 from step 3 into the equation for T^2 from step 4:
\(T^2 = (\frac{2 \pi R}{\sqrt{G M R^{(2-n)}}})^2\)
\(T^2 = \frac{4 \pi^2 R^2}{G M R^{(2-n)}}\),
Then, simplify to find the relationship between T and R:
\(T^2 = \frac{4 \pi^2}{G M} R^{(n+1)}\)
Hence, \(T \propto R^{[(n+1)/2]}\).
6Step 6: Find the correct answer
Comparing the expression found in step 5 with the given options, we can conclude that the correct answer is:
(A) \(R^{[(n+1) / 2]}\).
Key Concepts
Gravitational ForceCircular OrbitsTime Period of PlanetsInverse Power Law of Distance
Gravitational Force
Gravitational force is a fundamental force of nature that acts between two masses. Imagine it as a pulling force, keeping planets in their orbits and people grounded on Earth. It is always attractive, and it acts over vast distances in the universe. According to Newton's law of universal gravitation, this force is directly proportional to the masses of the objects and inversely proportional to the square of the distance between them. In mathematical terms, it's expressed as:
- \( F = \frac{G M m}{R^2} \)
Circular Orbits
Picture a planet moving around the sun in a path that is perfectly circular. A circular orbit means that the distance (or radius \( R \)) between the planet and the sun remains constant. Such an orbit provides a simplified scenario to analyze celestial mechanics. The centripetal force required to maintain this circular path is supplied by gravitational force. Therefore, the centripetal force equation becomes vital:
- \( \frac{m v^2}{R} = F \)
Time Period of Planets
The time period of a planet in orbit, often seen as \( T \), is the time it takes for a planet to complete one full orbit around the sun. Kepler’s third law traditionally relates this time period to the semi-major axis of the orbit (for circular paths, this is the radius \( R \)) with the relation:
- \( T^2 \propto R^3 \)
- \( T^2 = \frac{4 \pi^2}{G M} R^{(n+1)} \)
- Simplifying further, \( T \propto R^{[(n+1)/2]} \)
Inverse Power Law of Distance
The inverse power law of distance is an important concept when modifying gravitational laws. Normally, the gravitational force is inversely proportional to the square of the distance, represented as \( R^2 \). However, the problem introduces an alternate scenario where the force varies inversely as the \( n \)th power of the distance:
- \( F = \frac{G M m}{R^n} \)
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