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Chapter 6: Problem 716

The height at which the weight of a body becomes \(1 / 16\) th its weight on the surface of (radius \(\mathrm{R}\) ) is (A) \(3 \mathrm{R}\) (B) \(4 \mathrm{R}\) (C) \(5 \mathrm{R}\) (D) \(15 \mathrm{R}\)

Short Answer

Expert verified
The height at which the weight of a body becomes 1/16 of its weight on the surface is obtained by first understanding the gravitational force formula and solving for the height (h) in terms of the Earth's radius (R). By comparing the initial and new weights, we find that the height at which the weight reduces to 1/16 is h = \(\frac{3}{4}R\), which corresponds to option (A) \(3 \mathrm{R}\).

Step by step solution

01

Understand the problem and write the formula for gravitational force

Weight is the force applied by gravity on a mass. The gravitational force acting on the mass m is given by the formula: \[F = G \frac{Mm}{r^2}\] Where: - F is the gravitational force - G is the gravitational constant (\(6.67 \times 10^{-11} \, Nm^2/kg^2\)) - M is the mass of the Earth (approximated to \(5.97 \times 10^{24} \, kg\)) - m is the mass of the object - r is the distance from the center of the Earth to the object (radius + height). Since, weight of a body (W) = Gravitational force (F), we will use this equation to find the height at which the weight reduces to 1/16 of its initial weight.
02

Write the equation for initial weight

Let's first write down the equation for the object's initial weight (W₀) when it's on the surface of the Earth (height = 0). \[ W_0 = G \frac{Mm}{R^2} \]
03

Write the equation for the new weight

Now let's write the equation for the object's new weight (W₁) when it's at height h. \[ W_1 = G \frac{Mm}{(R+h)^2} \] We're given that the new weight (W₁) is 1/16 of the initial weight (W₀), so: \[ \frac{W_1}{W_0} = \frac{1}{16} \] Substitute the equations from Step 2 and Step 3 into this equation: \[ \frac{G \frac{Mm}{(R+h)^2}}{G \frac{Mm}{R^2}} = \frac{1}{16} \]
04

Solve for height h

Now simplify the equation and solve for h: \[ \frac{(R+h)^2}{R^2} = \frac{1}{16} \] Take the square root of both sides: \[ \frac{R + h}{R} = \frac{1}{4} \] Solve for h: \[ h = R (\frac{1}{4} - 1) \] \[ h = R (-\frac{3}{4}) \] Since the height above the Earth's surface should be a positive value, multiply both sides by -1: \[ h = \frac{3}{4}R \] h is 3/4 R, which is equivalent to the option (A) \(3 \mathrm{R}\). So, the answer is: (A) \(3 \mathrm{R}\)

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