Assume a light planet revolving around a very massive star in a circul...
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2R = F
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Assume a light planet revolving around a very massive star in a circul...
Gravitational force of attraction between the planet and the star:
The gravitational force of attraction between two objects is given by the equation:
F = G * (m1 * m2) / r^2
Where F is the force of attraction, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two objects.
According to the given information, the gravitational force of attraction between the planet and the star is proportional to R^(-3/2). This can be written as:
F ∝ R^(-3/2)
Using the equation for gravitational force, we can write this as:
G * (m1 * m2) / r^2 ∝ R^(-3/2)
We can simplify this equation by rearranging it:
R^(-3/2) ∝ (r^2) / (m1 * m2)
Since the planet is revolving around the star in a circular orbit, the distance between the planet and the star is equal to the radius of the orbit:
r = R
Substituting this into the equation, we get:
R^(-3/2) ∝ (R^2) / (m1 * m2)
Simplifying further, we can write:
R^(-3/2) ∝ R^2 / (m1 * m2)
Now, let's consider the relationship between the period of revolution (T) and the radius of the orbit (R).
Period of revolution and radius of the orbit:
The period of revolution (T) is the time it takes for the planet to complete one full revolution around the star. It is related to the radius of the orbit (R) by the equation:
T ∝ R^a
where 'a' is a constant that depends on the specific characteristics of the system.
From the given information, we know that the gravitational force of attraction is proportional to R^(-3/2). We can rewrite this as:
R^(-3/2) ∝ 1 / R^(3/2)
Comparing this with the expression T ∝ R^a, we can see that 'a' must be equal to -3/2 in order for the periods to be proportional to R^(-3/2).
Therefore, the correct answer is option 'D', T is proportional to R^(5/4).