Using the conservation laws, demonstrate that the total mechanical energy of a planet of mass m moving around the Sun along an ellipse depends only on its semi-major axis a. Find this energy as a function of a.

Options

A. `-(GM_(s)m)/(2a)`
B. `(GM_(s)m)/(2a)`
C. `(GM_(s)m)/(3a)`
D. `-(GM_(s)m)/(3a)`

Answer

To demonstrate that the total mechanical energy of a planet of mass \( m \) moving around the Sun along an ellipse depends only on its semi-major axis \( a \), we can use the conservation of mechanical energy in a gravitational field. Here's the step-by-step solution:

### Step 1: Define the Total Mechanical Energy
The total mechanical energy \( E \) of the planet in orbit is given by the sum of its kinetic energy \( K \) and potential energy \( U \):
\[ E = K + U \]
where the kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} m v^2 \]
and the gravitational potential energy \( U \) is given by:
\[ U = -\frac{G M m}{r} \]
Here, \( G \) is the gravitational constant, \( M \) is the mass of the Sun, \( m \) is the mass of the planet, and \( r \) is the distance from the planet to the Sun.

### Step 2: Use the Conservation of Angular Momentum
For a planet in elliptical orbit, the angular momentum \( L \) is conserved. The angular momentum is given by:
\[ L = m r^2 \omega \]
where \( \omega \) is the angular velocity. Since \( L \) is constant, we can express \( v \) in terms of \( r \) and \( \omega \):
\[ v = r \omega \]

### Step 3: Relate the Semi-Major Axis to the Orbit
In an elliptical orbit, the semi-major axis \( a \) is related to the maximum distance \( r_1 \) (aphelion) and minimum distance \( r_2 \) (perihelion) as follows:
\[ a = \frac{r_1 + r_2}{2} \]

### Step 4: Express Kinetic and Potential Energy in Terms of \( a \)
Using the vis-viva equation for elliptical orbits, we can express the velocity \( v \) at any point in the orbit as:
\[ v^2 = G M \left( \frac{2}{r} - \frac{1}{a} \right) \]
Substituting this into the expression for kinetic energy:
\[ K = \frac{1}{2} m \left( G M \left( \frac{2}{r} - \frac{1}{a} \right) \right) \]

### Step 5: Substitute into the Total Energy Equation
Now substituting \( K \) and \( U \) into the total energy equation:
\[ E = \frac{1}{2} m \left( G M \left( \frac{2}{r} - \frac{1}{a} \right) \right) - \frac{G M m}{r} \]
Combining the terms:
\[ E = \frac{G M m}{2a} - \frac{G M m}{r} \]

### Step 6: Simplify the Expression
For an elliptical orbit, the average distance \( r \) over one complete orbit can be approximated by the semi-major axis \( a \):
\[ E = -\frac{G M m}{2a} \]
This shows that the total mechanical energy \( E \) depends only on the semi-major axis \( a \).

### Final Result
Thus, the total mechanical energy of a planet moving in an elliptical orbit around the Sun is given by:
\[ E = -\frac{G M m}{2a} \]