NEET2026PhysicsChapterActual
A satellite A revolves around a planet of radius R in a circular orbit very close to its surface with a time period T . Another satellite B revolves around the same planet in a circular orbit at a height of 8R from the surface of the planet. The time period of satellite B will be:
Options
- A9T
- B16 2 T
- C27T
- D8T
Correct answer
C. 27T
Step-by-step solution
According to Kepler's third law, the square of the time period of a satellite is directly proportional to the cube of its orbital radius, T^2 r^3 . For satellite A , the orbital radius is r_A = R since it revolves very close to the surface of the planet. For satellite B , the height from the surface is 8R , so its orbital radius is r_B = R + 8R = 9R . Using the relation, we get T_B^2 T_A^2 = ( r_B r_A )^3 . Substituting the given values: T_B^2 T^2 = ( 9R R )^3 = 9^3 = 729 Taking the square root on both sides: T_B =