Notes : Newton's Law of Universal Gravitation Vector Form - Class 11 Physics

Newton's Law of Universal Gravitation (Vector Form and Displacement Vector Form)

Newton’s Law of Universal Gravitation states that every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Statement of Newton's Law of Gravitation

If two point masses \(m_1\) and \(m_2\) are separated by a distance \(r\), then the gravitational force between them is:

$$ F=G\frac{m_1m_2}{r^2} $$

Where:

• \(F\) = Gravitational force
• \(G\) = Universal gravitational constant
• \(m_1,m_2\) = Masses of the bodies
• \(r\) = Distance between their centers

Vector Form of Newton's Law of Gravitation

Let:

• \(\vec r_1\) = Position vector of mass \(m_1\)
• \(\vec r_2\) = Position vector of mass \(m_2\)

The displacement vector from \(m_1\) to \(m_2\) is:

$$ \vec r=\vec r_2-\vec r_1 $$

Magnitude of displacement vector:

$$ r=|\vec r| $$

Unit vector along the line joining the masses:

$$ \hat r=\frac{\vec r}{|\vec r|} $$

Therefore, the vector form of gravitational force is:

$$ \vec F=-G\frac{m_1m_2}{r^2}\hat r $$

Substituting

$$ \hat r=\frac{\vec r}{r} $$

we get

$$ \vec F = -G\frac{m_1m_2}{r^3}\vec r $$

Definition of Vector Form in Terms of Displacement Vector

The vector form of Newton's law of gravitation expresses the gravitational force in terms of both magnitude and direction using the displacement vector between two masses.

If

$$ \vec r=\vec r_2-\vec r_1 $$

then the gravitational force is

$$ \vec F= -G\frac{m_1m_2}{|\vec r|^3}\vec r $$

or

$$ \boxed{ \vec F= -G\frac{m_1m_2} {|\vec r_2-\vec r_1|^3} (\vec r_2-\vec r_1) } $$

Explanation of Symbols:

• \(\vec F\) = Gravitational force vector
• \(G\) = Universal gravitational constant
• \(m_1,m_2\) = Masses of the bodies
• \(\vec r_1,\vec r_2\) = Position vectors
• \(\vec r\) = Displacement vector
• \(|\vec r|\) = Magnitude of displacement vector

The negative sign indicates that gravitational force is attractive and acts opposite to the displacement vector.

Derivation of Vector Form in Terms of Displacement Vector

According to Newton's law of gravitation:

$$ F=G\frac{m_1m_2}{r^2} $$

To represent force as a vector, multiply by the unit vector \(\hat r\). Since the force is attractive, it acts opposite to \(\hat r\).

$$ \vec F=-G\frac{m_1m_2}{r^2}\hat r $$

Using

$$ \hat r=\frac{\vec r}{r} $$

Substituting:

$$ \vec F= -G\frac{m_1m_2}{r^2} \left(\frac{\vec r}{r}\right) $$ $$ \vec F= -G\frac{m_1m_2}{r^3}\vec r $$

Since

$$ \vec r=\vec r_2-\vec r_1 $$

Therefore,

$$ \boxed{ \vec F= -G\frac{m_1m_2} {|\vec r_2-\vec r_1|^3} (\vec r_2-\vec r_1) } $$

This is Newton’s law of gravitation expressed completely in terms of displacement vector.

Gravitational Forces Between Two Masses

Force on \(m_2\) due to \(m_1\)

$$ \vec F_{21} = \frac{Gm_1m_2}{r^2} \hat r_{21} $$

Force on \(m_1\) due to \(m_2\)

$$ \vec F_{12} = \frac{Gm_1m_2}{r^2} \hat r_{12} $$

Proof that Gravitational Forces are Equal and Opposite

$$ \hat r_{12} = -\hat r_{21} $$

Therefore,

$$ \vec F_{12} = -\vec F_{21} $$

Hence Newton’s law of gravitation obeys Newton’s Third Law of Motion.

Universal Gravitational Constant (G)

Value:

$$ G=6.67\times10^{-11}\;N\,m^2\,kg^{-2} $$

SI Unit:

$$ N\,m^2\,kg^{-2} $$

Dimensional Formula:

$$ [M^{-1}L^3T^{-2}] $$

Important Characteristics of Gravitational Force

• Gravitational force is always attractive.
• It acts along the line joining the centers of the masses.
• It is a central force.
• It obeys the inverse square law.
• It obeys Newton's Third Law.
• It depends only on masses and distance.

Solved Numerical

Question:

Calculate the gravitational force between two masses \(10\,kg\) and \(20\,kg\) separated by \(2\,m\). Take \(G=6.67\times10^{-11}\;N\,m^2\,kg^{-2}\).

Given:

$$ m_1=10\,kg $$ $$ m_2=20\,kg $$ $$ r=2\,m $$ $$ G=6.67\times10^{-11}\;N\,m^2\,kg^{-2} $$

Formula:

$$ F=G\frac{m_1m_2}{r^2} $$

Calculation:

$$ F= 6.67\times10^{-11} \times \frac{10\times20}{2^2} $$ $$ F= 6.67\times10^{-11} \times \frac{200}{4} $$ $$ F= 6.67\times10^{-11}\times50 $$ $$ F= 3.335\times10^{-9}\;N $$

Answer:

$$ \boxed{ F=3.335\times10^{-9}\;N } $$

Frequently Asked Questions (FAQ)

Q1. What is Newton's law of universal gravitation?

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

Q2. What is the vector form of Newton's law of gravitation?

$$ \vec F=-G\frac{m_1m_2}{r^2}\hat r $$

Q3. What is the displacement vector?

$$ \vec r=\vec r_2-\vec r_1 $$

Q4. Why is there a negative sign in the vector form?

It indicates that the gravitational force is attractive.

Q5. What is the SI unit of G?

$$ N\,m^2\,kg^{-2} $$

Q6. What is the dimensional formula of G?

$$ [M^{-1}L^3T^{-2}] $$

Quiz Questions

Multiple Choice Questions

1. The gravitational force between two masses is proportional to:

A. \(m_1+m_2\)
B. \(m_1-m_2\)
C. \(m_1m_2\)
D. \(\frac{m_1}{m_2}\)

Answer: C. \(m_1m_2\)

2. Gravitational force varies inversely as:

A. \(r\)
B. \(r^2\)
C. \(r^3\)
D. \(r^4\)

Answer: B. \(r^2\)

3. The SI unit of G is:

A. \(Nkg^{-1}\)
B. \(Nm^{-2}\)
C. \(Nm^2kg^{-2}\)
D. \(kgm^2s^{-2}\)

Answer: C. \(Nm^2kg^{-2}\)

4. Gravitational force is always:

A. Repulsive
B. Attractive
C. Zero
D. Variable

Answer: B. Attractive

Quick Revision Box

$$ F=G\frac{m_1m_2}{r^2} $$ $$ \vec F=-G\frac{m_1m_2}{r^2}\hat r $$ $$ \vec r=\vec r_2-\vec r_1 $$ $$ \vec F= -G\frac{m_1m_2} {|\vec r_2-\vec r_1|^3} (\vec r_2-\vec r_1) $$ $$ \vec F_{12}=-\vec F_{21} $$ $$ G=6.67\times10^{-11}\;N\,m^2\,kg^{-2} $$ $$ [M^{-1}L^3T^{-2}] $$