Problem 29
Question
The height at which the acceleration due to gravity decreases by \(36 \%\) of its value on the surface of the earth. (The radius of the earth is \(R\) ). (a) \(\frac{R}{6}\) (b) \(\frac{R}{4}\) (c) \(\frac{R}{2}\) (d) \(\frac{2}{3} R\)
Step-by-Step Solution
Verified Answer
d
1Step 1: Understanding Gravity’s Variation with Height
The acceleration due to gravity (abla gabla gdabla g = \frac{GM}{R^2} \frac{GM}{R^2} \frac{GM}{R^2} \frac{GM}{R^2}) is inversely proportional to the square of the distance from the center of the Earth, meaning it decreases with increasing height. It is related to the height above the surface by \( g_h = g \left( \frac{R}{R + h} \right)^2 \) where \( g_h \) is the gravity at height \( h \).
Key Concepts
Gravitational ForceVariation of Gravity with AltitudeInverse Square Law
Gravitational Force
Gravitational force is a fundamental force of nature that attracts two masses towards each other. This force happens because any object with mass exerts a force on other objects with mass.
The Earth, being massive, exerts a significant gravitational force that pulls objects towards its center. The formula for gravitational force is given by Newton's law of universal gravitation:
This is an important concept to grasp when studying how gravity changes with altitude.
The Earth, being massive, exerts a significant gravitational force that pulls objects towards its center. The formula for gravitational force is given by Newton's law of universal gravitation:
- Formula: \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \)
- Where \( F \) is the gravitational force, \( G \) is the universal gravitational constant, \( m_1 \) and \( m_2 \) are the masses involved, and \( r \) is the distance between the centers of the two masses.
This is an important concept to grasp when studying how gravity changes with altitude.
Variation of Gravity with Altitude
When you increase your altitude, say by climbing a mountain or flying in an airplane, the gravitational force exerted by the Earth decreases. Simply put, the further you are from the Earth's center, the less gravitational pull there is on you.
Gravity on the Earth's surface is strongest at sea level. As you ascend, the strength of gravity diminishes. The rate at which gravity decreases with altitude can be estimated using the formula:
For everyday altitudes like the height of Mount Everest, the change in gravitational acceleration is relatively small.
Gravity on the Earth's surface is strongest at sea level. As you ascend, the strength of gravity diminishes. The rate at which gravity decreases with altitude can be estimated using the formula:
- \( g_h = g \left( \frac{R}{R + h} \right)^2 \)
- Where \( g_h \) is the gravity at height \( h \), \( g \) is the standard gravity at the Earth's surface, and \( R \) is the Earth's radius.
For everyday altitudes like the height of Mount Everest, the change in gravitational acceleration is relatively small.
Inverse Square Law
The inverse square law is a pivotal principle in physics that describes how a physical quantity decreases with the square of the distance from the source. This concept heavily influences how gravity operates in our universe.
For gravity, this law states that the force decreases proportionally to the square of the distance from the Earth's center. When we talk about the inverse square law in terms of gravity:
Understanding the inverse square law helps in computing gravity's weakening effect as one moves away from the Earth's surface, providing significant insights into how our world works.
For gravity, this law states that the force decreases proportionally to the square of the distance from the Earth's center. When we talk about the inverse square law in terms of gravity:
- As the distance \( r \) increases, gravitational force \( F \) decreases as \( \frac{1}{r^2} \).
- For example, doubling the distance from the Earth's center will cause the gravitational force to decrease by a factor of four (\(2^2\)).
Understanding the inverse square law helps in computing gravity's weakening effect as one moves away from the Earth's surface, providing significant insights into how our world works.
Other exercises in this chapter
Problem 28
How much energy will be necessary for making a body of \(500 \mathrm{~kg}\) escape from the earth? \(\left(g=9.8 \mathrm{~ms}^{-2}\right.\), radius of earth \(\
View solution Problem 29
As there is no external torque, angular momentum will remain constant. When the tortoise moves from \(A\) to \(C\), figure, moment of inertia of the platform an
View solution Problem 29
Two identical satellites are at \(R\) and \(7 R\) away from earth surface, the wrong statement is ( \(R=\) Radius of earth) (a) ratio of total energy will be \(
View solution Problem 30
If the value of \(g\) acceleration due to gravity at earth surface is \(10 \mathrm{~ms}^{-2}\), its value in \(\mathrm{ms}^{-2}\) at the centre of the earth, wh
View solution