Assume that Earth is in circular orbit around the Sun with kinetic energy K and potential energy U, taken to be zero for infinite separation. Then, the relationship between K and U:

Options

A. is K=U
B. is K = −U
C. is K = U/2
D. is K = −U/2

Explanation

To find the relationship between the kinetic energy (K) and potential energy (U) of the Earth in a circular orbit around the Sun, we can follow these steps:

### Step 1: Understand the Forces Acting on Earth
The Earth is in a circular orbit around the Sun, which means that the gravitational force between the Earth and the Sun provides the necessary centripetal force for the Earth's circular motion.

### Step 2: Write the Expression for Centripetal Force
The required centripetal force (F_c) for an object moving in a circular path is given by:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( m \) is the mass of the Earth,
- \( v \) is the orbital velocity of the Earth,
- \( r \) is the radius of the orbit (distance from the Earth to the Sun).

### Step 3: Write the Expression for Gravitational Force
The gravitational force (F_g) between the Earth and the Sun is given by Newton's law of gravitation:
\[ F_g = \frac{GMm}{r^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Sun,
- \( m \) is the mass of the Earth,
- \( r \) is the distance between the centers of the Earth and the Sun.

### Step 4: Set the Centripetal Force Equal to the Gravitational Force
For the Earth to remain in a stable circular orbit, the centripetal force must equal the gravitational force:
\[ \frac{mv^2}{r} = \frac{GMm}{r^2} \]

### Step 5: Cancel Out Mass and Rearrange
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{v^2}{r} = \frac{GM}{r^2} \]
Rearranging gives:
\[ v^2 = \frac{GM}{r} \]

### Step 6: Calculate Kinetic Energy (K)
The kinetic energy (K) of the Earth in its orbit is given by:
\[ K = \frac{1}{2} mv^2 \]
Substituting \( v^2 \) from the previous step:
\[ K = \frac{1}{2} m \left(\frac{GM}{r}\right) = \frac{GMm}{2r} \]

### Step 7: Calculate Potential Energy (U)
The gravitational potential energy (U) of the Earth-Sun system is given by:
\[ U = -\frac{GMm}{r} \]

### Step 8: Relate Kinetic Energy to Potential Energy
From the expressions derived:
- Kinetic Energy: \( K = \frac{GMm}{2r} \)
- Potential Energy: \( U = -\frac{GMm}{r} \)

Now, we can express the kinetic energy in terms of potential energy:
\[ K = -\frac{1}{2} U \]

### Conclusion
Thus, the relationship between the kinetic energy (K) and potential energy (U) of the Earth in a circular orbit around the Sun is:
\[ K = -\frac{1}{2} U \]