A satellite is orbiting with areal velocity `v_(a)`. At what height form the surface of the earth, it is rotating, if the radius of earth is `R`?

Options

A. `(4v_(a)^(2))/(gR^(2))-R`
B. `(2v_(a)^(2))/(gR^(2))-R`
C. `(v_(a)^(2))/(gR^(2))-R`
D. `(v_(a)^(2))/(2gR^(2))-R`

Explanation

To solve the problem of finding the height \( h \) from the surface of the Earth at which a satellite is orbiting with areal velocity \( v_a \), we follow these steps:

### Step 1: Understand Areal Velocity
Areal velocity \( v_a \) is defined as the area swept out by the satellite in a given time. Mathematically, it can be expressed as:
\[ v_a = \frac{dA}{dt} \]
where \( dA \) is the area swept out in time \( dt \).

### Step 2: Relate Areal Velocity to Orbital Velocity
For a satellite in a circular orbit, the areal velocity can also be expressed in terms of the radius \( r \) of the orbit and the orbital velocity \( v \):
\[ v_a = \frac{1}{2} r v \]
where \( v \) is the orbital velocity of the satellite.

### Step 3: Express Orbital Velocity
The orbital velocity \( v \) of a satellite in orbit at a distance \( r \) from the center of the Earth can be given by:
\[ v = \sqrt{\frac{GM}{r}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

### Step 4: Substitute Orbital Velocity into Areal Velocity
Substituting the expression for \( v \) into the areal velocity formula:
\[ v_a = \frac{1}{2} r \sqrt{\frac{GM}{r}} = \frac{1}{2} \sqrt{GMr} \]

### Step 5: Rearranging for Radius
To find the radius \( r \) in terms of \( v_a \), we can rearrange the equation:
\[ v_a = \frac{1}{2} \sqrt{GMr} \]
Squaring both sides gives:
\[ v_a^2 = \frac{1}{4} GMr \]
Rearranging for \( r \):
\[ r = \frac{4v_a^2}{GM} \]

### Step 6: Relate Radius to Height
The radius \( r \) is the distance from the center of the Earth to the satellite. If \( R \) is the radius of the Earth, then:
\[ r = R + h \]
where \( h \) is the height above the Earth's surface. Substituting this into the equation gives:
\[ R + h = \frac{4v_a^2}{GM} \]

### Step 7: Solve for Height \( h \)
Rearranging this equation to solve for \( h \):
\[ h = \frac{4v_a^2}{GM} - R \]

### Final Expression
Thus, the height \( h \) from the surface of the Earth is given by:
\[ h = \frac{4v_a^2}{gR} - R \]
where \( g \) is the acceleration due to gravity at the surface of the Earth, and \( g = \frac{GM}{R^2} \).

### Summary
The height \( h \) from the surface of the Earth at which the satellite is orbiting is:
\[ h = \frac{4v_a^2}{gR} - R \]