An asteroid of mass `2 xx 10^(-4) M_e`, where `M_e` is the mass of the earth, revolves in a circular orbit around the sun at a distance that is twice earth's distance from the sun. Find the ratio of the kinetic energy of the asteroid to that of earth

Options

A. `0.9 xx 10^(-4) `
B. `1.6 xx 10^(-5)`
C. `3.6 xx 10^(-5)`
D. `1.0 xx 10^(-4)`

Explanation

To find the ratio of the kinetic energy of the asteroid to that of the Earth, we can follow these steps:

### Step 1: Understand the Kinetic Energy Formula
The kinetic energy (K) of an object in a gravitational field can be expressed as:
\[ K = \frac{G M m}{2R} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the central body (in this case, the Sun),
- \( m \) is the mass of the orbiting body (the asteroid or the Earth),
- \( R \) is the distance from the central body.

### Step 2: Define the Masses and Distances
Given:
- Mass of the asteroid, \( M_A = 2 \times 10^{-4} M_e \) (where \( M_e \) is the mass of the Earth).
- Distance of the asteroid from the Sun, \( R_A = 2 R_e \) (where \( R_e \) is the distance of the Earth from the Sun).

### Step 3: Calculate the Kinetic Energy of the Asteroid
Using the formula for kinetic energy:
\[ K_A = \frac{G M_s M_A}{2 R_A} \]
Substituting the values:
\[ K_A = \frac{G M_s (2 \times 10^{-4} M_e)}{2 (2 R_e)} \]
\[ K_A = \frac{G M_s (2 \times 10^{-4} M_e)}{4 R_e} \]
\[ K_A = \frac{G M_s M_e}{2 R_e} \times \frac{2 \times 10^{-4}}{4} \]
\[ K_A = \frac{G M_s M_e}{2 R_e} \times \frac{10^{-4}}{2} \]
\[ K_A = \frac{G M_s M_e}{2 R_e} \times 5 \times 10^{-5} \]

### Step 4: Calculate the Kinetic Energy of the Earth
For the Earth, the kinetic energy is:
\[ K_E = \frac{G M_s M_e}{2 R_e} \]

### Step 5: Find the Ratio of Kinetic Energies
Now, we can find the ratio of the kinetic energy of the asteroid to that of the Earth:
\[ \frac{K_A}{K_E} = \frac{\frac{G M_s M_e}{2 R_e} \times 5 \times 10^{-5}}{\frac{G M_s M_e}{2 R_e}} \]
The terms \( \frac{G M_s M_e}{2 R_e} \) cancel out:
\[ \frac{K_A}{K_E} = 5 \times 10^{-5} \]

### Final Answer
Thus, the ratio of the kinetic energy of the asteroid to that of the Earth is:
\[ \frac{K_A}{K_E} = 5 \times 10^{-5} \]

---