A point mass is orbiting a significant mass M lying at the focus of the elliptical orbit having major and minor axes given by 2 a and 2b respectively. Let r be the distance between the mas M and the end point of major or axis. Velocity of the particle can be given as

Options

A. `(ab)/r sqrt((GM)/(a^(3)))`
B. `(ab)/rsqrt((GM)/(b^(3)))`
C. `(ab)/(2r)sqrt((GM)/(r^(3)))`
D. `(2ab)/rsqrt((GM)/(((a+b)/2)^(3)))`

Explanation

To find the velocity of a point mass orbiting a significant mass \( M \) at the focus of an elliptical orbit with major and minor axes \( 2a \) and \( 2b \) respectively, we can use Kepler's laws and some properties of elliptical orbits.

### Step-by-Step Solution:

1. **Understanding the Orbit**:
The point mass is moving in an elliptical orbit around the mass \( M \). The semi-major axis is \( a \) and the semi-minor axis is \( b \). The distance \( r \) is the distance from the mass \( M \) to the point mass at the end of the major axis.

2. **Kepler's Second Law**:
According to Kepler's second law, the line joining the planet to the sun sweeps out equal areas in equal times. This can be expressed mathematically as:
\[ \frac{dA}{dt} = \frac{L}{2m} \]
where \( L \) is the angular momentum of the orbiting mass.

3. **Angular Momentum**:
The angular momentum \( L \) for a point mass \( m \) moving in a circular path can be expressed as:
\[ L = mvr \]
where \( v \) is the orbital speed and \( r \) is the radius of the orbit.

4. **Area of the Ellipse**:
The area \( A \) of the elliptical orbit is given by:
\[ A = \pi ab \]
Therefore, the rate of area swept out per unit time can be written as:
\[ \frac{dA}{dt} = \frac{\pi ab}{T} \]
where \( T \) is the period of the orbit.

5. **Relating Area and Angular Momentum**:
Setting the two expressions for \( \frac{dA}{dt} \) equal gives:
\[ \frac{\pi ab}{T} = \frac{L}{2m} \]

6. **Substituting for \( L \)**:
Substituting \( L = mvr \) into the equation gives:
\[ \frac{\pi ab}{T} = \frac{mvr}{2m} \implies \frac{\pi ab}{T} = \frac{vr}{2} \]

7. **Solving for Velocity \( v \)**:
Rearranging the equation to solve for \( v \):
\[ v = \frac{2\pi ab}{Tr} \]

8. **Using Kepler's Third Law**:
Kepler's third law states that:
\[ T^2 = \frac{4\pi^2 a^3}{GM} \]
Therefore, the period \( T \) can be expressed as:
\[ T = 2\pi \sqrt{\frac{a^3}{GM}} \]

9. **Substituting \( T \) back into the velocity equation**:
Substitute \( T \) into the velocity equation:
\[ v = \frac{2\pi ab}{\left(2\pi \sqrt{\frac{a^3}{GM}}\right)r} \]
Simplifying gives:
\[ v = \frac{ab \sqrt{GM}}{a^3} = \frac{ab \sqrt{GM}}{a^2} = \frac{b\sqrt{GM}}{a} \]

### Final Result:
The velocity of the point mass is given by:
\[ v = \frac{b \sqrt{GM}}{a} \]