Gravitational Field Intensity due to Ring | JEE, NEET & CBSE Board Class 11 Exams Notes

Gravitational Field Intensity due to Ring at point on the axis

Let us consider a ring of mass M, having radius ‘a’; the gravitational field at a distance x along its axis is found as follows:

Consider a small length element along the circumferential length of the ring which has a mass ‘dm’; the field intensity due to this length element is given by,

dE = G(dm/r²)

The vertical components of the fields cancel each other due to the symmetry of the ring, and only horizontal components survive and add up to,

\(\begin{array}{l}E = \int_{0}^{2\pi}\frac{Gdm}{r^{2}}cos\alpha\end{array} \)

Since

\(\begin{array}{l}cos\alpha =\frac{x}{\sqrt{{{x}^{2}}+{{a}^{2}}}}\end{array} \),

On substituting, we get

\(\begin{array}{l}E=\frac{GMx}{{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{{}^{3}/{}_{2}}}}\end{array} \)

Which is the formula for Gravitational Field Intensity Due to a Ring at a distance x from center of ring of radius a.

**Note : At Center of ring, x = 0, so E = 0. Hence Gravitational Field Intensity at center of a Ring is zero.

Important Question-Answers on Gravitational Field Intensity Due to a Ring

Q1: What is the Gravitational Field Intensity?

A: The gravitational field intensity at a point is the force experienced by a unit mass placed at that point in the gravitational field of a body.

Q2: What is the Formula for Gravitational Field Intensity Due to a Ring?

A: The gravitational field intensity EE at a point P on the axis of a ring of mass MM and radius RR, at a distance xx from the center, is given by:

\(\begin{array}{l}E=\frac{GMx}{{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{{}^{3}/{}_{2}}}}\end{array} \)

where:

• G is the universal gravitational constant,
• M is the mass of the ring,
• R is the radius of the ring,
• x is the distance of the point P from the center along the axis.

Q3: Why is the Gravitational Field at the Center of the Ring Zero?

A: At the center of the ring, the gravitational field due to each infinitesimal mass element of the ring cancels out symmetrically, leading to a net field intensity of zero.

Q4: How Does the Gravitational Field Vary Along the Axis?

A:

• At the center of the ring (x = 0), the field is zero.
• As x increases, the field initially increases, reaches a maximum, and then decreases asymptotically at large distances.

Q5: How Can This Concept Be Applied in Physics?

A: The concept of the gravitational field due to a ring is useful in astrophysics, planetary science, and engineering applications, including:

• Understanding the gravitational effects of planetary rings.
• Designing stable orbits in space missions.
• Theoretical studies of symmetrical mass distributions.

FAQs on Gravitational Field Intensity Due to a Ring

Q1: Is the Gravitational Field Due to a Ring Uniform?

A: No, the gravitational field varies depending on the location of the point along the axis and is not uniform.

Q2: What Happens to the Gravitational Field as x → ∞?

A: As the distance x increases, the field intensity decreases and approximates the field due to a point mass MM at large distances.

Q3: What is the Direction of the Gravitational Field at a Point on the Axis?

A: The gravitational field is directed towards the center of the ring along the axis.

MCQs on Gravitational Field Intensity Due to a Ring

Q1: The gravitational field intensity at the center of a uniform ring is:

A) Maximum
B) Zero
C) Equal to GM/R²
D) Infinite

Answer: B) Zero
Explanation: The field intensity at the center cancels due to symmetry.

Q2: The gravitational field intensity due to a ring along its axis is maximum at:

A) The center of the ring
B) Infinity
C) A certain distance from the center
D) On the ring itself

Answer: C) A certain distance from the center
Explanation: The field first increases, reaches a maximum, and then decreases with distance.

Q3: The formula for the gravitational field intensity due to a ring is derived using:

A) Newton’s laws of motion
B) Newton’s law of gravitation
C) Einstein’s relativity
D) Kepler’s laws

Answer: B) Newton’s law of gravitation
Explanation: The field intensity is derived based on the inverse-square law of gravitation.

Test Your Knowledge

Buy Complete Study Material

For detailed study materials, solved examples, and practice questions on JEE, NEET, and CBSE Class 11 Physics, visit Anand Classes Study Material.

Proprietor: NIRMAL ANAND Educations
Written by: Neeraj Anand
Published by: Anand Technical Publishers Under Anand Classes
Contact: +91-9463138669
Email: anandclasses1996@gmail.com