Planet Orbit Period and Radius Relationship

The problem describes a light planet orbiting a star in a circular path with radius R and period T. The gravitational force of attraction between the star and the planet is given to be proportional to $R^{-5/2}$. We need to find how the square of the period ($T^2$) is related to the radius (R) under this specific force law.

For a planet in a stable circular orbit around a star, the gravitational force provides the necessary centripetal force for the circular motion. Let the mass of the planet be $m$.

• Centripetal force ($F_c$) required for circular motion of radius R with velocity $v$ is given by:
• $$F_c = \frac{mv^2}{R}$$
• The orbital speed ($v$) of the planet in a circular orbit of radius R and period T is:
• $$v = \frac{2\pi R}{T}$$
• Substituting the expression for $v$ into the centripetal force formula:
• $$F_c = \frac{m}{R} \left(\frac{2\pi R}{T}\right)^2 = \frac{m}{R} \left(\frac{4\pi^2 R^2}{T^2}\right) = \frac{4\pi^2 m R}{T^2}$$

The gravitational force ($F_g$) between the star and the planet is given to be proportional to $R^{-5/2}$. We can write this as:

$$F_g = k R^{-5/2}$$

where $k$ is a constant of proportionality (which would depend on the masses of the star and planet, and the gravitational constant, but also incorporates the given unusual proportionality).

For the planet to stay in orbit, the gravitational force must equal the centripetal force:

$$F_g = F_c$$

Substituting the expressions for $F_g$ and $F_c$:

$$k R^{-5/2} = \frac{4\pi^2 m R}{T^2}$$

Now, we need to rearrange this equation to find the relationship between $T^2$ and R. Let's solve for $T^2$:

$$T^2 \cdot k R^{-5/2} = 4\pi^2 m R$$

$$T^2 = \frac{4\pi^2 m R}{k R^{-5/2}}$$

Using the rules of exponents ($R^a / R^b = R^{a-b}$ and $1/R^b = R^{-b}$), we can simplify the term involving R:

$$R / R^{-5/2} = R^1 \cdot R^{5/2} = R^{1 + 5/2} = R^{2/2 + 5/2} = R^{7/2}$$

So the equation for $T^2$ becomes:

$$T^2 = \left(\frac{4\pi^2 m}{k}\right) R^{7/2}$$

Since the term $\left(\frac{4\pi^2 m}{k}\right)$ is a constant (assuming the masses of the star and planet and the constant $k$ do not change), we can conclude that $T^2$ is proportional to $R^{7/2}$.

$$T^2 \propto R^{7/2}$$

Let's check the given options:

• Option 1: $T^2 \propto R^3$ (This is Kepler's Third Law for the standard inverse square gravitational force, which is $F \propto R^{-2}$, not $R^{-5/2}$)
• Option 2: $T^2 \propto R^{3/2}$
• Option 3: $T^2 \propto R^{7/2}$

Our derived relationship matches Option 3.