A hypothetical planet of mass m is moving along an elliptical path around sun of mass `M_s` under the influence of its gravitational pull. If the major axis is 2R, find the speed of the planet when it is at a distance of R from the sun.

Options

A. `sqrt((GM_s)/R)`
B. `sqrt(2GM_s)/R`
C. `2sqrt((GM_s)/R`
D. `m/Msqrt(GM_s)/R`

Answer

To solve the problem, we need to find the speed of a hypothetical planet of mass \( m \) when it is at a distance \( R \) from the Sun of mass \( M_s \). The major axis of the elliptical orbit is given as \( 2R \), which means the semi-major axis \( a \) is \( R \).

### Step-by-Step Solution:

1. **Identify the Semi-Major Axis**:
The semi-major axis \( a \) of the elliptical orbit is given by:
\[ a = \frac{2R}{2} = R \]

2. **Apply Kepler's Third Law**:
According to Kepler's Third Law, the square of the orbital period \( T \) of a planet is proportional to the cube of the semi-major axis \( a \):
\[ T^2 = \frac{4\pi^2 a^3}{G M_s} \]
Substituting \( a = R \):
\[ T^2 = \frac{4\pi^2 R^3}{G M_s} \]

3. **Calculate the Orbital Period \( T \)**:
Taking the square root of both sides to find \( T \):
\[ T = 2\pi \sqrt{\frac{R^3}{G M_s}} \]

4. **Determine the Speed \( v \)**:
The speed \( v \) of the planet at a distance \( R \) can be calculated using the formula for speed in terms of the distance traveled in one complete orbit divided by the period:
\[ v = \frac{\text{Circumference of orbit}}{T} \]
The circumference of the orbit (approximated as a circle at distance \( R \)) is:
\[ \text{Circumference} = 2\pi R \]
Thus, substituting for \( T \):
\[ v = \frac{2\pi R}{T} = \frac{2\pi R}{2\pi \sqrt{\frac{R^3}{G M_s}}} \]

5. **Simplify the Expression**:
The \( 2\pi \) cancels out:
\[ v = \frac{R}{\sqrt{\frac{R^3}{G M_s}}} \]
This can be simplified further:
\[ v = \frac{R \sqrt{G M_s}}{R^{3/2}} = \sqrt{\frac{G M_s}{R}} \]

### Final Result:
The speed of the planet when it is at a distance \( R \) from the Sun is:
\[ v = \sqrt{\frac{G M_s}{R}} \]