Question

If the time period \( T \) of a satellite revolving close to the earth is given as \( T = 2\pi R^a g^b \), then the value of \( a \) and \( b \) are respectively (where \( R \) is the radius of the earth):

• A. \( -\frac{1}{2} \) and \( -\frac{1}{2} \)
• B. \( \frac{1}{2} \) and \( -\frac{1}{2} \)
• C. \( \frac{1}{2} \) and \( \frac{1}{2} \)
• D. \( \frac{3}{2} \) and \( -\frac{1}{2} \)
• E. \( -\frac{1}{2} \) and \( \frac{1}{2} \)

Hint:

In problems involving the time period of a satellite, use the expression for gravitational force and ensure dimensional consistency to determine the exponents in the equation for \( T \).

Answer 1

The Correct Option is B

The time period \( T \) of a satellite revolving close to the earth is given by: \[ T = 2\pi R^a g^b \] where: - \( R \) is the radius of the earth,
- \( g \) is the acceleration due to gravity.
Now, the acceleration due to gravity \( g \) at a distance \( R \) from the center of the earth is given by the formula: \[ g = \frac{GM}{R^2} \] where: - \( G \) is the universal gravitational constant,
- \( M \) is the mass of the earth,
- \( R \) is the radius of the earth.
Substitute this expression for \( g \) into the equation for \( T \): \[ T = 2\pi R^a \left(\frac{GM}{R^2}\right)^b = 2\pi R^a \cdot \frac{(GM)^b}{R^{2b}} = 2\pi \cdot (GM)^b \cdot R^{a - 2b} \] For the time period to be dimensionally correct, the exponents of \( R \) and the constants must match the dimensions of time. By equating the powers of \( R \) and \( g \), we find that: - \( a - 2b = \frac{1}{2} \),
- \( b = -\frac{1}{2} \).
Thus, solving for \( a \) and \( b \), we get: \[ a = \frac{1}{2} \quad {and} \quad b = -\frac{1}{2}. \] Thus, the correct answer is option (B), \( \frac{1}{2} \) and \( -\frac{1}{2} \).