Problem 77 - A Complete Resource Book in Physics for JEE Main

Question

The height at which the acceleration due to gravity becomes \(\frac{g}{9}\) (where \(g\) = the acceleration due to gravity on the surface of the earth) in terms of \(R\), the radius of the earth is (A) \(\frac{R}{\sqrt{2}}\) (B) \(R / 2\) (C) \(\sqrt{2} R\) (D) \(2 R\)

Step-by-Step Solution

Answer

The height at which the acceleration due to gravity becomes \(\frac{g}{9}\) in terms of the radius of the earth is \(2R\). Thus, the answer is (D) \(2 R\).

Step 1: Set the equation

According to the law of gravitation, the acceleration due to gravity at a height \(h\) from the surface of the Earth is given by: \(g' = \frac{gR^2}{(R+h)^2}\). We are given that \(g'\) must be equal to \(\frac{g}{9}\), now substitute that into the equation.

Step 2: Solve for \(h\)

Setting up the equation: \(\frac{gR^2}{(R+h)^2} = \frac{g}{9}\). By cross multiplying and simplifying we get: \(9R^2 = R^2 + 2Rh + h^2\). Now we arrange for \(h\): \(8R^2 = 2Rh + h^2 = h(2R + h)\). Solving this equation for \(h\) gives us the value of \(h\) : \(h = 2R\).

Step 3: Check the result against the provided solution

Based on the solution we have, we find that answer (D) \(2 R\) matches our calculated height.

Key Concepts

Gravitational Law

The gravitational law is a fundamental principle in physics, explaining how two objects attract each other with a force proportional to their masses and inversely proportional to the square of the distance between them. This law is mathematically expressed by Newton's equation: \[F = G \frac{m_1 m_2}{r^2}\]where:

• \(F\) is the gravitational force between two objects.
• \(G\) is the gravitational constant \(6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2\).
• \(m_1\) and \(m_2\) are the masses of the two objects.
• \(r\) is the distance between the centers of the two masses.

Earth's gravitational force is crucial for calculations of gravity-related events, like determining the fall speed of objects or their weight. The acceleration due to gravity on Earth's surface, denoted by \(g\), is approximately \(9.8 \text{m/s}^2\). In our context, we analyze how gravity changes at different heights above Earth's surface.

Radius of the Earth

The radius of the Earth is a significant factor when calculating gravitational effects at various altitudes. Earth is not a perfect sphere, and its radius slightly varies from poles to the equator. However, for simplicity, an average radius of \(R = 6,371\text{ km}\) is often used for calculations.Understanding the radius helps in calculating changes in gravitational force as you ascend or descend relative to the Earth's surface.When solving problems regarding gravity, different formulas may involve the radius. For example, in the exercise solution, the change in gravity at height \(h\) uses the equation:\[g' = \frac{gR^2}{(R+h)^2}\]This equation reflects how gravity decreases with increasing distance \(R+h\) from the Earth's center.

Height and Gravity Relationship

Gravity varies with height, and understanding this relationship is essential for calculations in physics. As you move away from Earth's surface, the gravitational force decreases.The formula used in the original exercise captures this change:\[g' = \frac{gR^2}{(R+h)^2}\]where:

• \(g'\) is the acceleration due to gravity at height \(h\).
• \(g\) is the acceleration due to gravity on the surface.
• \(R\) is Earth's radius.
• \(h\) is the height above Earth's surface.

In the exercise, when \(g'\) is set to \(\frac{g}{9}\), it demonstrates how gravity weakens as height increases. Solving the equation yields \(h = 2R\), showing that at a height twice Earth's radius \(2R\), gravity is reduced to one-ninth of its surface value. This concept is important in understanding orbital dynamics and satellite positioning.