For the moon to cease to remain the earth’s satellite its orbital velocity has to increase by a factor of

To find the factor by which the moon's orbital velocity must increase so that it ceases to be the Earth's satellite, we need to compare its current orbital velocity with the escape velocity required to leave the Earth's gravitational pull.

The orbital velocity of a satellite (like the moon) revolving in a circular orbit of radius $r$ around the Earth of mass $M$ is given by the formula:
$v_o=\sqrt{\frac{GM}{r}}$
where $G$ is the universal gravitational constant.

For the moon to cease to remain Earth's satellite and escape its gravitational field, its velocity must be increased to the escape velocity ($v_e$) at that distance. The escape velocity is given by the formula:
$v_e=\sqrt{\frac{2GM}{r}}$

We can relate the escape velocity to the orbital velocity as follows:
$v_e=\sqrt{2}\sqrt{\frac{GM}{r}}=\sqrt{2}{v}_o$

This shows that the escape velocity is $\sqrt{2}$ times the orbital velocity. Therefore, for the moon to escape Earth's gravity, its orbital velocity has to increase by a factor of $\sqrt{2}$.

The correct option is √2.