Question

The height at which the weight of a body becomes one third of its weight on the surface of earth is (R is the radius of earth)

• A. $\frac{R}{3}$
• B. $0.732 R$
• C. $0.414 R$
• D. $\frac{R}{6}$
• E. $\frac{R}{10}$

Hint:

Logic Tip: Do not use the approximation formula $g_h = g(1 - 2h/R)$ unless $h$ is specifically stated to be very small compared to $R$ (typically $< 5%$). Since the weight changes by a massive factor (1/3), the exact formula must be used.

Answer 1

The Correct Option is B

Concept:
The acceleration due to gravity $g_h$ at a height $h$ above the Earth's surface is given by the exact formula: $$g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$$ where $g$ is the acceleration at the surface and $R$ is the Earth's radius. Since Weight $W = mg$, the weight scales identically to the local gravity.
Step 1: Set up the given condition.
We are given that the weight at height $h$ is one-third of the weight on the surface: $$W_h = \frac{1}{3}W$$ $$m \cdot g_h = \frac{1}{3}m \cdot g$$ $$g_h = \frac{g}{3}$$
Step 2: Equate and solve for h/R.
Substitute $g_h$ into the gravity formula: $$\frac{g}{3} = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$$ Cancel $g$ from both sides and take the reciprocal: $$3 = \left(1 + \frac{h}{R}\right)^2$$
Step 3: Solve the algebraic equation.
Take the square root of both sides: $$\sqrt{3} = 1 + \frac{h}{R}$$ We know that $\sqrt{3} \approx 1.732$. Substitute this value: $$1.732 = 1 + \frac{h}{R}$$ Subtract 1 from both sides: $$\frac{h}{R} = 0.732$$ $$h = 0.732 R$$