Question

A body is projected vertically from earth's surface of radius R with velocity equal to half the escape velocity. The maximum height reached by the body is

• A. \(\frac{R}{3}\)
• B. \(R\)
• C. \(\frac{R}{2}\)
• D. \(\frac{R}{4}\)

Hint:

Use energy conservation with the gravitational potential energy and the value of v squared from half the escape speed.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The escape speed is \(v_e = \sqrt{2GM/R}\). A launch speed of \(v = v_e/2\) means \(v^2 = \dfrac{v_e^2}{4} = \dfrac{GM}{2R}\).

Step 2: Energy conservation.
At the top the speed is zero. The loss of kinetic energy equals the gain of gravitational potential energy:
\[ \frac{1}{2}mv^2 = GMm\left(\frac{1}{R} - \frac{1}{R + h}\right) \]

Step 3: Solve.
Put \(v^2 = \dfrac{GM}{2R}\):
\[ \frac{1}{4R} = \frac{1}{R} - \frac{1}{R + h} \Rightarrow \frac{1}{R + h} = \frac{3}{4R} \Rightarrow R + h = \frac{4R}{3} \]
\[ h = \frac{R}{3} \]

Step 4: Check the options.
\(h = R\) would need launch speed \(\sqrt{GM/R}\), which is \(v_e/\sqrt{2}\), and the other options do not come from the equation.

Final Answer:
The maximum height is \(\dfrac{R}{3}\), option (A). \[ \boxed{\frac{R}{3}} \]