A satellite is orbiting around the earth with a period T. If the earth suddenly shrinks to half its radius without change in mass, the period of revolution of the satellite will be

The correct option is T.

Step-by-Step Explanation:

The time period of revolution of a satellite orbiting a planet of mass $M$ in an orbit of radius $r$ (distance from the center of the planet to the satellite) is given by Kepler's Third Law of planetary motion:

$T=2\pi \sqrt{\frac{r^3}{GM}}$

where:
- $T$ is the orbital period of the satellite.
- $r$ is the orbital radius of the satellite, measured from the center of the Earth.
- $G$ is the universal gravitational constant.
- $M$ is the mass of the Earth.

From this relation, we can see that the orbital period $T$ of the satellite depends only on the mass of the Earth $M$ and the orbital radius $r$. It is completely independent of the physical radius of the Earth itself.

When the Earth suddenly shrinks to half its physical radius without any change in its mass:
1. The mass $M$ of the Earth remains unchanged.
2. The orbital radius $r$ of the satellite (which is measured from the center of the Earth) remains constant.

Since both $M$ and $r$ do not change, the period of revolution of the satellite remains unaffected.

Therefore, the period of revolution of the satellite will remain T.