Understanding the gravitational field due to a uniform spherical shell is crucial for students preparing for JEE, NEET, and CBSE Board Class 11 Exams. This concept is extensively used in physics, especially in celestial mechanics and astrophysics.
You can buy complete study material at Anand Classes.
Gravitational Field Intensity due to Uniform Spherical Shell
A uniform spherical shell is a symmetrical mass distribution where the mass is uniformly spread over the surface of a sphere. The gravitational field varies at different positions:
Consider a thin uniform spherical shell of radius ‘R’, and mass ‘M’ situated in a space. A 3D object divides space into 3 parts:
Inside the spherical shell.
On the surface of the spherical shell.
Outside the spherical shell.
Our problem is to find out the value of gravitational field intensity in all these 3 regions.
Outside the Spherical Shell (r > R)
Consider a unit test mass at a point ‘P’, which is at a distance ‘r’ from the centre of the spherical shell. Draw an imaginary spherical shell such that point ‘P’ lies on the surface of it.
In this case the shell behaves as if all its mass is concentrated at its center. As we know, the gravitational field intensity at a point depends only upon the source mass and the distance of the point from the source mass. We can say that within the imaginary sphere, the source mass is M, and the distance of separation is ‘r’ from the center. From this, we get
E = -GM/r2
E ∝ -1/r2
Therefore, Gravitational Field Intensity decreases as r increases.
On the Surface of Spherical Shell (r = R)
Consider a unit test mass at a point ‘P’, which lies on the surface of the spherical shell at a distance ‘r’ from the centre of the spherical shell, then r = R. As discussed above, the gravitational field intensity on the surface of the spherical shell is given by,
E = -GM/R2
E = g = Constant
Hence, At the Spherical Shell (r = R), the gravitational field intensity is constant and is equal to acceleration due to gravity.
Inside the Spherical Shell (r < R)
If we consider a point inside the spherical shell, the entire mass of the shell lies above the point. Draw an imaginary spherical shell about point ‘P’; the source mass within this imaginary sphere will be zero.
We know that if the source mass is equal to zero, the gravitational field intensity is also equal to zero. Therefore,
E = 0
Hence, Inside the Spherical Shell (r < R), the gravitational field intensity is equal to zero.
Position of Point ‘P’
Gravitational Field Intensity
Inside the spherical shell (r < R)
E = 0
On the surface of the spherical shell (r = R)
E = -GM/R2
Outside the spherical shell ( r > R)
E = -GM/r2
Frequently Asked Questions (FAQs)
Q1: Why is the gravitational field inside a uniform spherical shell zero?
Answer: According to Newton’s Shell Theorem, for any point inside a uniform spherical shell, the gravitational forces due to all mass elements cancel out, leading to a net gravitational field of zero.
Q2: How does the gravitational field change as we move away from the shell?
Answer: The gravitational field behaves similarly to that of a point mass. Outside the shell, the field decreases as 1/r², and inside, it remains zero.
Q3: What is the significance of this concept in real-life applications?
Answer: This principle is fundamental in planetary science, satellite motion, and astrophysics, helping scientists understand the gravitational influence of celestial bodies.
Multiple-Choice Questions (MCQs)
Q1: What is the gravitational field inside a uniform spherical shell?
A)E = -GM/R2 B)E = -GM/r2 C) Zero D) None of the above
Correct Answer: C) Zero Explanation: Newton’s Shell Theorem states that the net gravitational field inside a uniform spherical shell is zero because the contributions from different mass elements cancel out.
Q2: How does the gravitational field outside a uniform spherical shell behave?
A) Increases with distance B) Decreases as C) Remains constant D) Becomes zero
Correct Answer: B) Decreases as Explanation: Outside the shell, the shell acts as a point mass located at its center, and its gravitational field follows the inverse square law.
Q3: If the radius of a spherical shell is doubled, how does the gravitational field at its surface change?
A) Becomes half B) Becomes one-fourth C) Remains the same D) Doubles
Correct Answer: B) Becomes one-fourth Explanation: The gravitational field at the surface is given by . If R is doubled, the field decreases by a factor of 4.
Gravitational Field Intensity Due to Spherical Shell Quiz
const questions = [
{ q: “What is the gravitational field intensity inside a hollow spherical shell?”, options: [“Zero”, “Maximum”, “Depends on mass”, “Equal to surface value”], answer: 0, explanation: “Inside a hollow spherical shell, the gravitational field intensity is zero due to symmetry.” },
{ q: “What is the gravitational field intensity outside a spherical shell?”, options: [“Same as a point mass”, “Zero”, “Infinite”, “Depends on shell thickness”], answer: 0, explanation: “Outside the shell, the field behaves as if all mass were concentrated at the center.” },
{ q: “What happens to the gravitational field intensity as we move inside a spherical shell?”, options: [“Increases”, “Decreases”, “Remains zero”, “Oscillates”], answer: 2, explanation: “Inside a uniform shell, the gravitational field is always zero.” },
{ q: “What is the formula for gravitational field intensity outside a spherical shell?”, options: [“g = Gm/r²”, “g = Gm/r”, “g = Gm/r³”, “g = Gm×r”], answer: 0, explanation: “Outside a shell, field intensity follows Newton’s law: g = Gm/r².” },
{ q: “What happens to the gravitational field intensity at the surface of a spherical shell?”, options: [“It is maximum”, “It is zero”, “It is half the outer value”, “It follows inverse square law”], answer: 3, explanation: “At the surface, the field obeys the inverse square law with distance.” },
{ q: “Which law helps determine the gravitational field inside a spherical shell?”, options: [“Newton’s First Law”, “Gauss’s Law”, “Kepler’s Law”, “Hooke’s Law”], answer: 1, explanation: “Gauss’s Law explains why the field inside a shell is zero.” },
{ q: “What factor affects gravitational field intensity outside a spherical shell?”, options: [“Only radius”, “Only mass”, “Both mass and radius”, “Only thickness of shell”], answer: 2, explanation: “Outside the shell, field intensity depends on both mass and distance from center.” },
{ q: “What happens to gravitational field intensity if the mass of the shell doubles?”, options: [“Doubles”, “Halves”, “Remains the same”, “Becomes zero”], answer: 0, explanation: “Since g = Gm/r², doubling mass doubles field intensity outside the shell.” },
{ q: “How does gravitational field intensity behave at the center of a spherical shell?”, options: [“Maximum”, “Zero”, “Equal to surface value”, “Infinite”], answer: 1, explanation: “At the exact center, the field is zero as forces from all directions cancel out.” },
{ q: “What happens to gravitational field intensity if the radius of the shell increases, keeping mass constant?”, options: [“Increases”, “Decreases”, “Remains the same”, “Becomes infinite”], answer: 1, explanation: “Larger radius spreads mass over a larger area, reducing field intensity.” }
];function loadQuiz() {
let quizContainer = document.getElementById(“quiz”);
quizContainer.innerHTML = “”;
questions.forEach((q, index) => {
let questionHTML = `
Understanding the gravitational field due to a uniform spherical shell is essential for solving advanced problems in physics. This topic is fundamental for JEE, NEET, and CBSE Board Class 11 students.
For comprehensive study materials and expert guidance, visit Anand Classes.
