Q.The planet Mars has two moons, phobos and delmos.

The escape speed from a planet’s surface is the minimum speed needed for an object to leave its gravitational pull forever. The formula is $v_e = \sqrt{\frac{2GM}{R}}$, where $M$ is the planet’s mass and $R$ its radius. The key insight: if the mean density $\rho$ is the same as Earth’s, then mass and radius are linked through $\rho = \frac{M}{\frac{4}{3}\pi R^3}$, so $R \propto M^{1/3}$. That lets us find how $v_e$ changes when only $M$ changes, without needing numerical values.

1. Write the escape speed formula.

   For Earth: $v_e = \sqrt{\frac{2GM}{R}}$. For the other planet: $v_e' = \sqrt{\frac{2G M'}{R'}}$, where $M' = 8M$ and $\rho' = \rho$.

2. Relate radius to mass using constant density.

   Since $\rho = \frac{M}{\frac{4}{3}\pi R^3}$ is the same for both,

So the planet’s radius is twice Earth’s.

3. Substitute into the escape speed ratio.

Therefore $v_e' = 2 \times 11.2\ \mathrm{km\,s^{-1}} = 22.4\ \mathrm{km\,s^{-1}}$. …

The escape velocity from a celestial body is the minimum speed needed for an object to break free from its gravitational pull without further propulsion. The formula is derived from energy conservation: kinetic energy at launch equals the work done against gravity to infinity. For a planet of mass $M$ and radius $R$, the escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$, where $G$ is the universal gravitational constant.

The key insight: escape velocity scales as the square root of mass over radius. So if we know how mass and escape velocity compare between two planets, we can solve for the unknown radius ratio.

Let’s denote Earth’s mass as $M_e$ and radius as $R_e = R$. The planet’s mass is $M_p = \frac{1}{10} M_e$. Its escape velocity $v_{e,p}$ is given as $\frac{1}{\sqrt{2}}$ times Earth’s escape velocity $v_{e,e}$.

1. Write the escape velocity for Earth:

2. Write the escape velocity for the planet:

3. The problem states:

Substitute the expressions:

4. Square both sides to remove square roots: $\frac{2G \cdot \frac{1}{10} M_e}{R_p} = \frac{1}{2} \cdot \frac{2GM_e}{R}$ …

We start with the concept: Earth’s rotation is determined by its angular momentum, which is conserved if no external torque acts. When the radius changes, the mass distribution changes, altering the moment of inertia. Since angular momentum $L = I \omega$ stays constant, the angular speed $\omega$ must adjust inversely to $I$. The period $T = 2\pi / \omega$ then changes proportionally to $I$.

1. Moment of inertia of a sphere For a solid sphere of mass $M$ and radius $R$, the moment of inertia about its axis is $I = \frac{2}{5} M R^2$. If the radius becomes $x$ times the present value, the new radius is $R' = xR$, so the new moment of inertia is

2. Conservation of angular momentum No external torque acts (we assume the change happens internally or gradually), so

Substituting $I' = I x^2$ gives

3. Relation between period and angular speed The period $T = 2\pi / \omega$. The new period is $$$$

The question tests your knowledge of Kepler’s laws of planetary motion and a famous experimental result. Kepler’s laws are:

1. Law of Ellipses – planets move in ellipses with the Sun at one focus.
2. Law of Equal Areas – a line joining a planet to the Sun sweeps out equal areas in equal times.
3. Law of Harmonies – the square of the orbital period is proportional to the cube of the semi-major axis.

Statement (D) refers to Cavendish’s experiment, which measured the gravitational constant $G$. That is a true historical fact. So the false statement must be among (A)–(C). Let’s check each carefully.

1. Statement (A): “All planets move in elliptical orbits with sun at one of the foci.”

   This is exactly Kepler’s first law. It is true.

   Pitfall: Some might think orbits are perfectly circular, but Kepler showed they are ellipses (circles are a special case of ellipses, but the Sun is at a focus, not the center).

2. Statement (B): “The square root of time period of revolution of planet is proportional to the cube of semi major axis of the ellipse.”

   Kepler’s third law says: $T^2 \propto a^3$, where $T$ is the period and $a$ is the semi-major axis.

   Taking square roots: $\sqrt{T^2} = T \propto a^{3/2}$. That is not “square root of time period proportional to cube of semi-major axis.” The statement says $\sqrt{T} \propto a^3$, which is false. …