3.3 Gravitational and Electric Forces

3.3 Gravitational and Electric Forces: Understanding the Fundamentals

Gravitational Force: The Universal Attraction

Gravity is the force by which one object with mass attracts another object with mass. From keeping the planets in orbit to shaping the vast structures of galaxies, gravitational force governs many phenomena in the universe.

What is Gravitational Force?

Gravitational force is always attractive and acts at a distance, making it a long-range force. Unlike contact forces, gravity does not require objects to touch for its effects to be felt.

Equation for Gravitational Force

The force of gravity on an object near the Earth’s surface is calculated using:

$$F_g = m \cdot g$$

Where:

• 
  $F_{\mathrm{g<span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;\; font-size:\; 16px;">is\; the\; force\; of\; gravity\; (N),</span>}}$
• 
  $m$

  is the mass of the object (kg),

• 
  $g$

  is the acceleration due to gravity (9.8 m/s2 on Earth).

This equation shows that the greater the mass of an object, the stronger the gravitational force acting on it.

Key Insight: The gravitational force becomes significant only when the masses of objects are very large, such as planets or stars.

Newton’s Universal Law of Gravitation

Newton’s Law explains that every particle in the universe attracts every other particle. The gravitational force is:

$$F_g = G \frac{{m_1 \cdot m_2}}{{r^2}}$$

​​

Where:

• 
  $F_g$

  ​ is the gravitational force (N),

• 
  $G$

  is the gravitational constant (6.67×

  $6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$
• 
  $m_1, m_2$

  ​ are the masses of the objects (kg),

• 
  $r$

  is the distance between the centers of the masses (m).

Key Relationships in Newton’s Law

1. Mass and Gravitational Force:

   • The force is directly proportional to the product of the masses
     $m_1\cdot m_{\mathrm{2<span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;\; font-size:\; 16px;"></span><span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;\; font-size:\; 16px;"></span>}}$
   • Larger masses result in stronger gravitational attraction.

2. Distance and Gravitational Force:

   • The force is inversely proportional to the square of the distance
     $1/r^2$
   • Doubling the distance decreases the force to one-fourth.

Gravitational Constant $G$

The small value of

$G$

explains why gravitational forces are typically weak unless very large masses are involved. For instance, while a pencil exerts a gravitational pull on you, it is negligible compared to Earth’s pull due to the pencil’s small mass.

Applications of Gravitational Force

• Orbits: Gravity keeps planets orbiting the Sun and moons orbiting planets.
• Galaxies: Gravity holds stars and planetary systems together within galaxies.
• Comets: Gravity governs the motion of comets around stars.
• Large-Scale Universe: Gravitational forces shape the structure and behavior of galaxy clusters.

Practice Problems

Problem 1:

Two objects with masses of 10 kg and 5 kg are 3 meters apart. What is the gravitational force between them?

• a)
  $0.22 \, \text{N}$
• b)
  $2.2 \, \text{N}$
• c)
  $22 \, \text{N}$
• d)
  $220 \, \text{N}$

Answer: a)

$0.22 \, \text{N}$

Explanation:

$$F=G\frac{m_1\cdot m_2}{r^2}=\frac{(6.67\times 10^{-11})(10)(5)}{3^2}=0.22\text{N}$$

Problem 2:

A planet with a mass of

$5 \times 10^{24} \, \text{kg}$

 is

$1.5 \times 10^{11} \, \text{m}$

 away from a star with a mass of

$2 \times 10^{30} \, \text{kg}$

. What is the gravitational force between them?

• a)
  $2.22 \times 10^{11} \, \text{N}$
• b)
  $2.22 \times 10^{13} \, \text{N}$
• c)
  $2.22 \times 10^{15} \, \text{N}$
• d)
  $2.22 \times 10^{17} \, \text{N}$

Answer: d)

$2.22 \times 10^{17} \, \text{N}$

Explanation:

$$F=G\frac{m_1\cdot m_2}{r^2}=\frac{(6.67\times 10^{-11})(5\times 10^{24})(2\times 10^{30})}{(1.5\times 10^{11}{)}^2}=2.22\times 10^{17}\text{N}$$

Problem 3:

A satellite with a mass of

$500 \, \text{kg}$

500kg is located

$1000 \, \text{km}$

1000km from the center of Earth (

$5.97 \times 10^{24} \, \text{kg}$

5.97×1024kg). What is the gravitational force acting on the satellite?

• a)
  $1.96 \times 10^{4} \, \text{N}$
• b)
  $3.92 \times 10^{4} \, \text{N}$
• c)
  $7.84 \times 10^{4} \, \text{N}$
• d)
  $1.57 \times 10^{5} \, \text{N}$

Answer: c)

$7.84 \times 10^{4} \, \text{N}$

Explanation:

$$F = G \frac{{m_1 \cdot m_2}}{{r^2}} = \frac{{(6.67 \times 10^{-11})(500)(5.97 \times 10^{24})}}{{(1000 \times 10^3)^2}} = 7.84 \times 10^{4} \, \text{N}$$