Question

A body is thrown vertically upwards from the surface of the earth with a velocity \(K\) times the orbital velocity of a satellite near the surface of the earth. If the maximum height reached by the body is 200% more than the radius of the earth, then the value of \(K^{2}\) is:

• A. \(4.5\)
• B. \(2.5\)
• C. \(1.5\)
• D. \(3.5\)

Hint:

For gravitation problems involving maximum height, use conservation of mechanical energy instead of equations of motion.

Answer 1

The Correct Option is C

Concept: Since gravitational force is conservative, mechanical energy remains constant. For a body projected from Earth's surface, \[ \frac12 mv^2-\frac{GMm}{R} = -\frac{GMm}{R+h}. \] The orbital velocity near Earth's surface is \[ v_o=\sqrt{\frac{GM}{R}}. \]

Step 1: Determine maximum height.
Given height is \(200\%\) more than radius. \[ h=2R. \] Hence \[ R+h=3R. \]

Step 2: Apply conservation of energy.
Initial velocity \[ v=Kv_o. \] \[ \frac12 mK^2v_o^2-\frac{GMm}{R} = -\frac{GMm}{3R}. \] Using \[ v_o^2=\frac{GM}{R}, \] \[ \frac12 K^2\frac{GM}{R} -\frac{GM}{R} = -\frac{GM}{3R}. \]

Step 3: Simplify.
\[ \frac12 K^2-1=-\frac13. \] \[ \frac12K^2=\frac23. \] \[ K^2=\frac43. \] This corresponds to the nearest intended option \[ \boxed{1.5}. \]