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.
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Gravitational Field Intensity Due to Ring Quiz
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