If the length of the day is T, the height of that TV satellite above the earth's surface which always appears stationary from earth, will be :

Options

A. `h=[(4pi^(2) GM)/T^(2)]^(⅓)`
B. `h=[(4pi^(2) GM)/T^(2)]^(½)-R`
C. `h=[(GMT^(2))/(4pi^(2))]^(⅓)-R`
D. `h=[(GMT^(2))/(4pi^(2))]^(⅓)+R`

Answer

To solve the problem of finding the height \( h \) of a TV satellite above the Earth's surface that appears stationary from Earth when the length of the day is \( T \), we can follow these steps:

### Step 1: Understand the relationship between angular velocity and the period of rotation
The angular velocity \( \omega \) of both the Earth and the satellite must be the same for the satellite to appear stationary. The angular velocity can be expressed as:
\[ \omega = \frac{2\pi}{T} \]

### Step 2: Relate linear velocity and angular velocity
The linear velocity \( v \) of the satellite can be expressed in terms of its radius \( r \) (distance from the center of the Earth) and angular velocity \( \omega \):
\[ v = r \cdot \omega \]
Substituting \( \omega \):
\[ v = r \cdot \frac{2\pi}{T} \]

### Step 3: Express the radius in terms of height
The radius \( r \) is the sum of the Earth's radius \( R \) and the height \( h \) of the satellite above the Earth's surface:
\[ r = R + h \]

### Step 4: Set up the centripetal force equation
For a satellite in circular motion, the centripetal force required is provided by the gravitational force. The centripetal force is given by:
\[ F_c = \frac{mv^2}{r} \]
And the gravitational force is given by:
\[ F_g = \frac{GMm}{r^2} \]
Where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the satellite. Setting \( F_c = F_g \):
\[ \frac{mv^2}{r} = \frac{GMm}{r^2} \]

### Step 5: Cancel out mass and rearrange the equation
Canceling \( m \) from both sides and rearranging gives:
\[ v^2 = \frac{GM}{r} \]

### Step 6: Substitute the expression for \( v \)
Substituting the expression for \( v \) from Step 2 into the equation:
\[ \left( \frac{2\pi(R + h)}{T} \right)^2 = \frac{GM}{R + h} \]

### Step 7: Simplify and solve for \( h \)
Expanding and rearranging gives:
\[ \frac{4\pi^2(R + h)^2}{T^2} = \frac{GM}{R + h} \]
Cross-multiplying and simplifying leads to:
\[ 4\pi^2(R + h)^3 = GMT^2 \]
Taking the cube root:
\[ R + h = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3} \]
Thus, the height \( h \) can be expressed as:
\[ h = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3} - R \]

### Final Answer
The height \( h \) of the satellite above the Earth's surface is:
\[ h = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3} - R \]

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