Question

The geostationary satellite is revolving round the earth of mean radius \(R\). If \(g\) is the acceleration due to gravity at earth's surface, \(\omega\) is the angular speed of rotation of earth about its own axis and \(r\) is the radius of orbit of a satellite, then the correct relation between \(r\), \(\omega\), \(g\) and \(R\) is

• A. \(r^{2}=\frac{gR^{2}}{w^{3}}\)
• B. \(R^{2}=\frac{r^{3}w^{2}}{g}\)
• C. \(R^{3}=\frac{r^{2}w}{g^{2}}\)
• D. \(R=\sqrt{\frac{rw^{3}}{g}}\)

Hint:

Equate gravitational force to the centripetal force and replace GM with gR squared.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A geostationary satellite has the same angular speed \(\omega\) as the earth. Gravity supplies the centripetal force.

Step 2: Calculation
\(\dfrac{GMm}{r^{2}}=m\omega^{2}r\), so \(GM=\omega^{2}r^{3}\).

Step 3: Use g
At the earth's surface \(g=\dfrac{GM}{R^{2}}\), so \(GM=gR^{2}\). Hence \(gR^{2}=\omega^{2}r^{3}\), i.e. \(R^{2}=\dfrac{r^{3}\omega^{2}}{g}\).

Final Answer:
The relation is R squared equals r cubed omega squared divided by g. \[ \boxed{\text{(B) }R^{2}=\dfrac{r^{3}\omega^{2}}{g}} \]