Problem 716 - JEE Physics

Question

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) \(15 \mathrm{R}\)

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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). \[ W_0 = G \frac{Mm}{R^2} \]

Step 3: 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: \[ \frac{G \frac{Mm}{(R+h)^2}}{G \frac{Mm}{R^2}} = \frac{1}{16} \]

Step 4: 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) \(3 \mathrm{R}\)

Key Concepts

Gravitational Constant

The gravitational constant, denoted as \(G\), is a pivotal part of understanding gravitational forces. It is a universal constant that appears in Newton's law of universal gravitation. This constant allows us to calculate the gravitational force between two masses.
Here are some key points to remember about the gravitational constant:

• Symbol: \(G\)
• Value: \(6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\)
• Role: Connects mass and distance with gravitational force.

By using \(G\), we ensure that calculations of gravitational forces are consistent and comparable no matter where in the universe they are applied. Understanding \(G\) is crucial as it provides the foundation for predicting how objects will behave under the influence of gravity.

Mass of the Earth

The Mass of the Earth is an essential component in calculating the gravitational force exerted on objects. Along with the gravitational constant, it helps us to determine how much gravitational pull the Earth will exert on an object of a particular mass.
Here’s what you need to know about Earth's mass:

• Symbol: \(M\)
• Approximate Value: \(5.97 \times 10^{24} \, \text{kg}\)
• Importance: Allows for the calculation of gravitational force exerted by Earth.

The immense size of Earth's mass means it plays a major role in producing significant gravitational force. Without knowing Earth's mass, it would be challenging to make accurate predictions about gravitational interactions with objects.

Distance from Center of Earth

The distance from the center of the Earth is vital in determining an object's weight or gravitational force experienced. This distance includes both the radius of the Earth and any additional height above Earth's surface.
Key considerations about this concept:

• Symbolized as \(r\) when considering a point above Earth’s surface.
• Combines Earth's radius with the object's height.
• Used in the formula \(r = R + h\), where \(R\) is Earth's radius and \(h\) is the height above the surface.

The gravitational force \(F\) is inversely proportional to \(r^2\), meaning as the distance increases, the force decreases. This concept is paramount when understanding how weight changes with altitude.

Weight Reduction by Fraction

The weight reduction by fraction when an object is above Earth's surface illustrates how gravitational force diminishes with distance. When a problem specifies that weight is reduced by a fraction, such as \(\frac{1}{16}\), it indicates how much weaker the gravitational pull becomes.
Understanding this involves:

• Knowing that weight is proportional to gravitational force.
• Using the equation: if \(W_1\) is \(\frac{1}{16}\) the weight at Earth’s surface \(W_0\), then \(\frac{W_1}{W_0} = \frac{1}{16}\).
• Finding the new distance \((R + h)\) by solving the equation \(\frac{G\frac{Mm}{(R+h)^2}}{G\frac{Mm}{R^2}} = \frac{1}{16}\).

Understanding weight reduction helps us appreciate how gravity weakens with elevation, explaining why objects weigh less at higher altitudes above Earth's surface.