Question

Statement I: The escape velocity for a planet at a point is $V_e$. A body is projected at that point horizontally with a speed V. This particle moves as satellite around the planet for any value of V satisfies the condition $\frac{V_e}{4} < V < V_e$.

Statement II: The escape velocity of body at any point is always equal to $\sqrt{2}$ times to orbital velocity at that point in planets gravitational field.

• A. Statement I is incorrect but Statement II is correct.
• B. Both Statement I and Statement II are correct.
• C. Both Statement I and Statement II are incorrect.
• D. Statement I is correct but Statement II is incorrect.

Answer 1

Answer: (A)

Statement I is incorrect but Statement II is correct.

Answer 2

Statement I: For a body to move as a satellite (in a circular or elliptical orbit) around a planet, its velocity $V$ at a point must satisfy $V_{orb} \le V < V_e$, where $V_{orb}$ is the orbital velocity and $V_e$ is the escape velocity at that point. The relationship between escape velocity and orbital velocity for a circular orbit is $V_e = \sqrt{2} V_{orb}$, which implies $V_{orb} = \frac{V_e}{\sqrt{2}}$. Thus, for satellite motion, the condition is $\frac{V_e}{\sqrt{2}} \le V < V_e$. The statement claims satellite motion occurs for $\frac{V_e}{4} < V < V_e$. Since $\frac{V_e}{4} < \frac{V_e}{\sqrt{2}}$, this range includes velocities less than the orbital velocity. For $V < V_{orb}$, the body will not maintain orbit and will fall back to the planet. Therefore, Statement I is incorrect.

Statement II: The escape velocity $V_e$ from a point at a distance $r$ from the center of a planet of mass $M$ is given by $V_e = \sqrt{\frac{2GM}{r}}$. The orbital velocity $V_{orb}$ for a circular orbit at the same distance $r$ is given by $V_{orb} = \sqrt{\frac{GM}{r}}$. Comparing these two, we find $V_e = \sqrt{2} \times \sqrt{\frac{GM}{r}} = \sqrt{2} V_{orb}$. This relationship holds true for any point in the planet's gravitational field. Therefore, Statement II is correct.