Question

A body is projected vertically upwards from the surface of the earth with a velocity equal to half the escape velocity. If \(R\) is the radius of the earth, the maximum height attained by the body from the surface of the earth is

• A. \(\dfrac{R}{6}\)
• B. \(\dfrac{R}{3}\)
• C. \(\dfrac{2R}{3}\)
• D. \(R\)

Hint:

Use energy conservation for variable gravity problems: \(\frac{1}{2}mv^2 - \frac{GMm}{r} = \text{constant}\).

Answer 1

The Correct Option is B

Step 1: Escape velocity relation.
Escape velocity:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
Given initial velocity:
\[ u = \frac{v_e}{2} \]
Step 2: Use energy conservation.
Total energy at surface:
\[ E = \frac{1}{2}mu^2 - \frac{GMm}{R} \]
At maximum height \(R+h\), velocity becomes zero:
\[ E = -\frac{GMm}{R+h} \]
Step 3: Substitute \(u = \frac{v_e}{2}\).
\[ \frac{1}{2}m\left(\frac{v_e}{2}\right)^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]
\[ \frac{1}{8}mv_e^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]
But \(v_e^2 = \frac{2GM}{R}\), so:
\[ \frac{1}{8}m\cdot \frac{2GM}{R} - \frac{GMm}{R} = -\frac{GMm}{R+h} \]
\[ \left(\frac{1}{4}-1\right)\frac{GMm}{R} = -\frac{GMm}{R+h} \]
\[ -\frac{3}{4}\frac{GMm}{R} = -\frac{GMm}{R+h} \]
Step 4: Solve for height.
Cancel \(-GMm\):
\[ \frac{3}{4R} = \frac{1}{R+h} \]
\[ R+h = \frac{4R}{3} \Rightarrow h = \frac{4R}{3}-R = \frac{R}{3} \]
Final Answer: \[ \boxed{\dfrac{R}{3}} \]