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Chapter 9: Problem 20

The depth at which the effective value of acceleration due to gravity is \(\frac{g}{4}\) is \((R=\) radius of the earth) [MP PET 2003] (a) \(R\) (b) \(\frac{3 R}{4}\) (c) \(\frac{R}{2}\) (d) \(\frac{R}{4}\)

Short Answer

Expert verified
The correct answer is (b) \(\frac{3 R}{4}\).

Step by step solution

01

Understand the Variation Formula

We use the formula for the variation of gravity with the radius of the Earth, which is given by \(g鈥� = g (1 - \frac{d}{R})\) where \(g鈥橽) is the effective acceleration due to gravity at a depth \(d\), \(g\) is the acceleration due to gravity at the surface, and \(R\) is the radius of the Earth.
02

Set up the Equation and Solve for d

Given that \(g鈥� = \frac{g}{4}\), we substitute this into our formula and solve for \(d\). The equation becomes \(\frac{g}{4} = g (1 - \frac{d}{R})\). Multiply both sides by 4 to get \(g = 4g (1 - \frac{d}{R})\). After cancelling out \(g\) on both sides, the equation simplifies to \(1 = 4 - 4 \frac{d}{R}\). Solving for \(d\) finally yields, \(d = \frac{3R}{4}\)
03

Identify the correct answer choice

After solving for \(d\) and observing that it is \(\frac{3R}{4}\), we can conclusively ascertain that the correct choice of this multiple choice question is (b) \(\frac{3 R}{4}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration due to gravity
The acceleration due to gravity, commonly denoted by the symbol \( g \), is the rate at which an object accelerates when falling freely towards the surface of the Earth. It is approximately \( 9.8 \, m/s^2 \) at the surface. This means that if you drop an object from a certain height, its velocity will increase by about 9.8 meters per second for each second it falls.
  • Surface Value: The standard value of \( g \) applies at the Earth's surface where it is strongest.
  • Factors affecting \( g \): Gravity varies based on several factors, including the object's distance from the Earth鈥檚 center, which means it changes with altitude and depth.
Gravity is an essential concept in physics as it affects the motion of objects at or near the Earth. Understanding this concept helps in solving problems like the variation of gravity with depth or height.
Radius of the Earth
The radius of the Earth (\( R \)) is a measure of the distance from the Earth's center to its surface. It is a critical factor when calculating gravitational variations. The approximate average radius of the Earth is about 6371 kilometers. However, this value may vary slightly because the Earth is not a perfect sphere; it is slightly flattened at the poles and bulging at the equator.
  • Importance: The radius is crucial when calculating how gravity changes with depth below or height above the Earth's surface.
  • Applications in Formulas: The radius is used in various formulas, including the gravitational variation formula which calculates how \( g \) changes with depth \( d \).
Understanding the Earth's radius helps us interpret physical phenomena and solve exercises relating to gravitational changes effectively.
Gravitational variation formula
The gravitational variation formula describes how gravity changes with depth inside the Earth. According to this formula, as you move deeper below the Earth's surface, the effective gravity decreases.The formula is expressed as:\[g' = g \left(1 - \frac{d}{R}\right)\]where \( g' \) is the acceleration due to gravity at depth \( d \), \( g \) is the gravitational acceleration at the surface, and \( R \) is the Earth鈥檚 radius.
  • Interpretation: The term \( \frac{d}{R} \) represents the fraction of the Earth's radius you are beneath the surface, signifying how much gravity is reduced by at that depth.
  • Calculations: By knowing the effective gravity at a specific depth, you can rearrange this formula to find that depth, as shown in the solution to the exercise.
This formula is vital as it allows us to calculate changes in gravity for different depths, enhancing our understanding of Earth's internal gravitational field.

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Most popular questions from this chapter

The acceleration of a body due to the attraction of the earth (radius \(R\) ) at a distance \(2 R\) from the surface of the earth is \((g=\) acceleration due to gravity at the surface of the earth) (a) \(\frac{g}{9}\) (b) \(\frac{g}{3}\) (c) \(\frac{g}{4}\) (d) \(g\)

The distance of Neptune and Saturn from sun are nearly \(10^{13}\) and \(10^{12}\) meters respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio [NCERT 1975; CBSE PMT 1994; MP PET (a) \(\sqrt{10}\) (b) 100 (c) \(10 \sqrt{10}\) (d) \(1 / \sqrt{10}\)

In planetary motion the areal velocity of position vector of a planet depends on angular velocity \((\omega)\) and the distance of the planet from sun \((r)\). If so the correct relation for areal velocity is (a) \(\frac{d A}{d t} \propto \omega r\) (b) \(\frac{d A}{d t} \propto \omega^{2} r\) (c) \(\frac{d A}{d t} \propto \omega r^{2}\) (d) \(\frac{d A}{d t} \propto \sqrt{\omega r}\)

A satellite is launched into a circular orbit of radius ' \(R\) ' around earth while a second satellite is launched into an orbit of radius \(1.02 R\). The percentage difference in the time periods of the two satellites is [EAMCET 2003] (a) \(0.7\) (b) \(1.0\) (c) \(1.5\) (d) 3

Weight of a body of mass \(m\) decreases by \(1 \%\) when it is raised to height \(h\) above the earth's surface. If the body is taken to a depth h in a mine, change in its weight is \(\quad\) [KCET 2003; MP PMT 2003] (a) \(2 \%\) decrease (b) \(0.5 \%\) decrease (c) \(1 \%\) increase (d) \(0.5 \%\) increase
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