मराठी

Newton's Universal Law of Gravitation

Advertisements

Topics

  • Introduction
  • History/Origin
  • Definition: Universal Law of Gravitation
  • Formula: Universal Law of Gravitation
  • Key Points: Newton's Universal Law of Gravitation
  • Characteristics
  • Relationship to the Acceleration of the Moon
  • Generalisation to Force
  • Force Due to the Collection of Masses
  • Special Cases for Extended Objects
  • Significance
  • Example 1
  • Example 2
  • Real-Life Examples
Maharashtra State Board: Class 10, 11

Definition: Universal Law of Gravitation

"Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them."

Maharashtra State Board: Class 10, 11

Formula: Universal Law of Gravitation

The gravitational force of attraction (F) between two bodies of mass m1 and m2 separated by a distance r is:

\[\mathbf{F} = \mathbf{G}\frac{m_1 m_2}{r^2}\]
  • F: Gravitational Force of attraction (in Newtons, N).

  • \[m_1, m_2\]: Masses of the two objects (in kilograms, kg).

  • r (or d in the first part): Distance between the two objects (in meters, m).

  • G: The constant of proportionality, called the Universal gravitational constant.

    • Value in SI units: \[G=6.67\times10^{-11}\mathrm{N}\cdot\mathrm{m}^2/\mathrm{kg}^2\]

    • Dimensions: \[[G]=[\mathrm{L}^3\mathrm{M}^{-1}\mathrm{T}^{-2}]\]

Maharashtra State Board: Class 10

Key Points: Newton's Universal Law of Gravitation

  • Every object attracts every other with a gravitational force.
  • Force increases with mass — more mass means a stronger pull.
  • Force decreases with distance — doubling the distance halves the force.
  • A force acts along the line joining the centres (or centres of mass) of the two bodies.
Maharashtra State Board: Class 11

Introduction

Newton's Universal Law of Gravitation is one of the most important laws in physics. Every object in the Universe attracts every other object with a definite force. This force depends on two things: the masses of the objects and the distance between them. The law explains why objects fall to Earth and why planets move around the Sun. This law was developed by Newton after studying the work of famous scientists like Kepler and Galileo. Understanding this law helps us understand how gravity works everywhere in the universe.

Maharashtra State Board: Class 11

History/Origin

  • 1564-1642: Galileo discovered that heavy and light objects fall to Earth with the same acceleration when released from the same height.
  • 1571-1630: Johannes Kepler, a German astronomer and mathematician, worked as a helper to Tycho Brahe in Prague starting in 1600. After Brahe's sudden death in 1601, Kepler became the Royal mathematician.
  • Kepler's Contribution: Kepler used observations of planetary positions made by Brahe to discover the laws of planetary motion. His work was later used by Newton in postulating the law of gravitation.
  • 1665: Newton studied the motion of the Moon around the Earth. He knew the Moon completes one revolution about the Earth in 27.3 days and is at a distance of 3.85 × 10⁵ km.
  • Newton's Achievement: Newton combined the ideas of Galileo and Kepler with his own laws of motion to formulate the Universal Law of Gravitation.
Maharashtra State Board: Class 11

Characteristics

  • Nature of Force: It is an attractive force.
  • Action-Reaction Pair: The forces exerted by the two bodies on each other have the same magnitude but opposite directions (obeying Newton's Third Law).
  • Line of Action: The force always acts along the line joining the centres (or centres of mass) of the two bodies.
  • Universality: The law applies to all material objects in the universe.
  • Impact of Mass: If the mass of one object is doubled, the force between the two objects doubles.
  • Impact of Distance: If the distance between the two objects is doubled, the force decreases by a factor of 4 (due to the inverse square relationship).
Maharashtra State Board: Class 11

Relationship to Acceleration of the Moon

1. Moon's Centripetal Acceleration: Newton studied the Moon's motion around the Earth, which is almost circular. The centripetal force required is F = mrω², leading to an acceleration of a = rω².

  • Using the period T (27.3 days) and distance r (\[3.85 \times 10^5\] km), the Moon's acceleration (\[a_{moon}\]) towards the Earth was calculated:
    \[a = r \left(\frac{2\pi}{T}\right)^2 \approx 0.0027 \, \text{m/s}^2\]

2. Comparison of Accelerations: This acceleration (0.0027 m/s²) is much smaller than the acceleration of objects near the Earth's surface (aobject ≈ 9.8 m/s²).

3. Inverse Square Relationship: By comparing the ratio of accelerations and the ratio of distances:

\[\frac{a_{object}}{a_{moon}} \approx 3600 \quad \text{and} \quad \frac{\text{distance of moon}}{\text{distance of object}} \approx 60\]

It was found that:

\[\frac{a_{object}}{a_{moon}} = \left[\frac{\text{distance of moon}}{\text{distance of object}}\right]^2\]

4. Conclusion: Newton concluded that the acceleration of an object towards the Earth is inversely proportional to the square of its distance from the center of the Earth (\[\mathbf{a}\propto1/\mathbf{r}^2\]).

Maharashtra State Board: Class 11

Generalization to Force

  1. Force on an Object: Since F = ma, the force exerted by the Earth on an object of mass m at a distance r is F ∝ m/r².
  2. Newton's Third Law: The object also exerts an equal and opposite force on the Earth (FE ∝ M/r²).
  3. Final Generalization: By combining these, Newton concluded that the gravitational force between the Earth (Mass M) and an object (Mass m) is F ∝ Mm/r².
  4. Universal Law: This was generalized to the force between any two objects (m1 and m2) as F ∝ m1m2/r².
Maharashtra State Board: Class 11

Force Due to Collection of Masses (Superposition Principle)

  • For a collection of point masses, the force on any one mass is the vector sum of the gravitational forces exerted by all the other point masses.
  • For n particles, the force on the ith mass (\[\vec F_i\]) is:
    \[F_i=\sum_{
    \begin{array}
    {l}j=1 \\
    j\neq i
    \end{array}}^nF_{ij}\]
Maharashtra State Board: Class 11

Special Cases for Extended Objects

The source provides two special cases for gravitational force involving extended objects:

  1. Force inside a hollow spherical shell: The gravitational force of attraction due to a hollow, thin spherical shell of uniform density, on a point mass situated inside it is zero.
  2. Force outside a hollow shell or solid sphere: The gravitational force between a hollow spherical shell or solid sphere of uniform density and a point mass situated outside is just as if the entire mass is concentrated at the center of the shell or sphere.
Maharashtra State Board: Class 11

Significance

  • It led to an explanation of terrestrial gravitation (why things fall on Earth).
  • It explains Kepler’s laws of planetary motion.
  • It provides the reason behind the observed motion of planets around the Sun.
  • It explains the force that keeps the Moon in its circular orbit around the Earth.
Maharashtra State Board: Class 11

Example 1

Problem: The gravitational force between two bodies is 1 N. If the distance between them is doubled, what will be the gravitational force between them?

Solution:

Let the masses of the two bodies be m₁ and m₂, and the distance between them be r.
F₁ = \[\frac{Gm_1m_2}{r^2}\]
When distance is doubled (new distance = 2r): F₂ = \[\frac{Gm_1m_2}{(2r)^2}=\frac{Gm_1m_2}{4r^2}\].
\[\frac{F_1}{F_2}=\frac{\frac{Gm_1m_2}{r^2}}{\frac{Gm_1m_2}{4r^2}}=4\]
F₂ = \[\frac {F_1}{4}\]
Since F₁ = 1 N:

F₂ = \[\frac {1}{4}\] = 0.25 N

Answer: When the distance is doubled, the gravitational force becomes one-fourth (0.25 N).

Maharashtra State Board: Class 11

Example 2

Problem: Three particles A, B, and C each having mass m are kept along a straight line with AB = BC = l. A fourth particle D is kept on the perpendicular bisector of AC at a distance l from B. Determine the gravitational force on D.

Solution:

First, we need to find the distances AD and CD:

CD = AD = \[\sqrt{(AB)^2+(BD)^2}=\sqrt{l^2+l^2}=\sqrt{2l^2}=\sqrt{2}\mathbf{l}\]

Force due to A on D:
\[F_{DA}=\frac{Gmm}{(\sqrt{2}l)^2}=\frac{Gm^2}{2l^2}\]

Force due to C on D:
\[\begin{aligned}
F_{DC}=\frac{Gmm}{(\sqrt{2}l)^2}=\frac{Gm^2}{2l^2}
\end{aligned}\]

Force due to B on D:
\[F_{DB}=\frac{Gmm}{(l)^2}=\frac{Gm^2}{l^2}\]

Let the unit vector along the horizontal direction (AC) be î and along the vertical direction (BD) be ĵ.

Horizontal Component (along î direction):

  • The force \[\vec F_{DB}\] is purely vertical, so its horizontal component is zero (cos 90° = 0).
  • The triangle ABD is a right isosceles triangle, so the angle ∠DAB = 45°.
  • FDA and FDC have equal horizontal components, but they are in opposite directions (canceling each other out).
  • \[\text{Net Horizontal Force} = F_{DC} \cos 45^\circ - F_{DA} \cos 45^\circ = 0\]

Vertical Component (along ĵ direction):

From A: \[\frac{Gm^2}{2l^2}\times\frac{1}{\sqrt{2}}=\frac{Gm^2}{2\sqrt{2}l^2}\]

From B: \[\frac{Gm^{2}}{2l^{2}}\times\frac{1}{\sqrt{2}}=\frac{Gm^{2}}{2\sqrt{2}l^{2}}\]

From C: \[\frac{Gm^2}{l^2}\]

Net Vertical Force:
= \[\frac{Gm^{2}}{2\sqrt{2}l^{2}}+\frac{Gm^{2}}{l^{2}}+\frac{Gm^{2}}{2\sqrt{2}l^{2}}\]

= \[\frac{Gm^2}{l^2}\left[\frac{1}{2\sqrt{2}}+1+\frac{1}{2\sqrt{2}}\right]\]

= \[\frac{Gm^2}{l^2}\left[\frac{2}{2\sqrt{2}}+1\right]=\frac{Gm^2}{l^2}\left[\frac{1}{\sqrt{2}}+1\right]\]

Final Answer:

The net gravitational force on D is \[{\frac{Gm^{2}}{l^{2}}}\left(1+{\frac{1}{\sqrt{2}}}\right)\] directed along DB (toward B).

Maharashtra State Board: Class 11

Real-Life Examples

  • Tides: The gravitational force exerted by the Moon and the Sun causes the ocean tides on Earth.
  • Satellite Launch: This force is used to calculate the required escape velocity for launching rockets and satellites into Earth's orbit.
  • The Solar System: Gravitation is the force that holds the planets, asteroids, and comets in their orbits around the Sun, structuring the entire solar system.
  • Walking/Standing: It is the force that keeps us grounded and gives us weight.

Test Yourself

Video Tutorials

We have provided more than 1 series of video tutorials for some topics to help you get a better understanding of the topic.

Series 1


Series 2


Shaalaa.com | Universal law of gravitation

Shaalaa.com


Next video


Shaalaa.com


Universal law of gravitation [00:17:59]
S
Advertisements
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×