Gravitation Notes

Gravitation

Introduction

• Objects in motion and the concept of force.
• Force is required to change the speed or direction of an object's motion.
• Objects dropped from a height fall towards the Earth.
• Planets revolve around the Sun, and the Moon revolves around the Earth.
• A force is responsible for these phenomena: gravitational force.
• This chapter covers gravitation, the universal law of gravitation, motion of objects under gravitational force, weight variation, and conditions for objects to float in liquids.

Gravitation

• The Moon revolves around the Earth.
• Objects thrown upwards fall downwards after reaching a certain height.
• Newton's observation: an apple falling from a tree.
• Question: If Earth can attract an apple, can it also attract the Moon?
• Conjecture: The same force is responsible for both.
• Argument: The Moon falls towards the Earth at each point of its orbit instead of going in a straight line; hence, it is attracted by the Earth.
• Activity 9.1: Stone tied to a thread whirled around.
  • Observation: The stone moves in a circular path with a certain speed and changes direction at every point.
  • Change in direction implies change in velocity or acceleration.
  • The force causing this acceleration and keeping the body moving along the circular path is directed towards the center: centripetal force.

Universal Law of Gravitation

• Every object in the universe attracts every other object.

• The force is proportional to the product of their masses.

• The force is inversely proportional to the square of the distance between them.

• The force acts along the line joining the centers of the two objects.

• Tangent to a circle: A straight line that meets the circle at one and only one point.

• The motion of the Moon around the Earth is due to the centripetal force provided by the Earth's attraction.

• If there were no such force, the Moon would move in a uniform straight line.

• A falling apple is attracted towards the Earth; the apple also attracts the Earth.

• According to the third law of motion, the apple does attract the earth. But according to the second law of motion, for a given force, acceleration is inversely proportional to the mass of an object $(Eq. 8.4)$.

• The mass of the apple is negligibly small compared to the Earth, so we don't see the Earth moving towards the apple.

• The same argument applies to why the Earth does not move towards the Moon.

• In the solar system, planets revolve around the Sun, indicating a force between them.

• Newton concluded that all objects in the universe attract each other: gravitational force.

• Mathematical representation:

  • Let two objects A and B have masses $M$ and $m$, respectively, and be separated by a distance $d$. Let $F$ be the force of attraction between them.
  • $F \propto M \times m$ $(9.1)$
  • $F \propto \frac{1}{d^2}$ $(9.2)$
  • Combining equations (9.1) and (9.2):
    • $F \propto \frac{M \times m}{d^2}$ $(9.3)$
    • $F = G \frac{M \times m}{d^2}$ $(9.4)$
    • Where $G$ is the universal gravitation constant.
  • $F \times d^2 = G M \times m$ $(9.5)$

• The SI unit of $G$ is $N m^2 kg^{-2}$.

• Henry Cavendish found the value of $G$ using a sensitive balance.

• The accepted value of $G$ is $6.673 \times 10^{-11} N m^2 kg^{-2}$.

• The law is universal, applicable to all bodies, big or small, celestial or terrestrial.

Inverse-square

• If $d$ is bigger by a factor of 6, $F$ becomes $\frac{1}{36}$ times smaller.
• Example 9.1: Calculating the force exerted by the Earth on the Moon.
  • Mass of the Earth, $M = 6 \times 10^{24} kg$
  • Mass of the Moon, $m = 7.4 \times 10^{22} kg$
  • Distance between the Earth and the Moon, $d = 3.84 \times 10^5 km = 3.84 \times 10^8 m$
  • $G = 6.7 \times 10^{-11} N m^2 kg^{-2}$
  • $F = \frac{G \times M \times m}{d^2} = \frac{6.7 \times 10^{-11} N m^2 kg^{-2} \times 6 \times 10^{24} kg \times 7.4 \times 10^{22} kg}{(3.84 \times 10^8 m)^2} = 2.02 \times 10^{20} N$
  • The force exerted by the Earth on the Moon is $2.02 \times 10^{20} N$.

Importance of the Universal Law of Gravitation

• Explains several phenomena:
  • The force that binds us to the Earth.
  • The motion of the Moon around the Earth.
  • The motion of planets around the Sun.
  • Tides due to the Moon and the Sun.

Free Fall

• Activity 9.2: Throwing a stone upwards.
• The Earth attracts objects towards it due to gravitational force.
• Objects falling towards the Earth under this force alone are in free fall.
• During free fall, there is no change in the direction of motion.
• The magnitude of velocity changes due to Earth's attraction.
• Any change in velocity involves acceleration.
• Acceleration due to the Earth's gravitational force: acceleration due to gravity ($g$).
• The unit of $g$ is $m s^{-2}$.
• From the second law of motion, force is the product of mass and acceleration.
  • $F = m g$ $(9.6)$
  • From equations (9.4) and (9.6):
    • $mg = G \frac{M m}{d^2}$
    • $g = G \frac{M}{d^2}$ $(9.7)$
  • Where $M$ is the mass of the Earth, and $d$ is the distance between the object and the Earth.
• For objects on or near the surface of the Earth, $d = R$ (radius of the Earth).
  • $g = G \frac{M}{R^2}$ $(9.9)$

To Calculate the Value of g

• $G = 6.7 \times 10^{-11} N m^2 kg^{-2}$
• $M = 6 \times 10^{24} kg$
• $R = 6.4 \times 10^6 m$
  • $g = \frac{G M}{R^2} = \frac{6.7 \times 10^{-11} N m^2 kg^{-2} \times 6 \times 10^{24} kg}{(6.4 \times 10^6 m)^2} = 9.8 m s^{-2}$
• The value of acceleration due to gravity of the Earth, $g = 9.8 m s^{-2}$.

Motion of Objects Under the Influence of Gravitational Force of the Earth

• Activity 9.3: Dropping a sheet of paper and a stone from a height.
• Paper reaches the ground later than the stone due to air resistance.
• Air resistance affects the paper more than the stone.
• In a glass jar from which air has been sucked out, the paper and the stone would fall at the same rate.
• Acceleration experienced by an object during free fall is independent of its mass.
• Galileo's experiment at the Leaning Tower of Pisa.
• All equations for uniformly accelerated motion are valid with $a$ replaced by $g$.
  • $v = u + at$ $(9.10)$
  • $s = ut + \frac{1}{2} at^2$ $(9.11)$
  • $v^2 = u^2 + 2as$ $(9.12)$
• Take $a$ as positive when it is in the direction of motion and negative when it opposes the motion.
• Example 9.2: A car falls off a ledge and drops to the ground in 0.5 s. Let $g = 10 m s^{-2}$.
  • (i) What is its speed on striking the ground?
  • (ii) What is its average speed during the 0.5 s?
  • (iii) How high is the ledge from the ground?
  • Solution:
    • Time, $t = 0.5 s$
    • Initial velocity, $u = 0 m s^{-1}$
    • Acceleration due to gravity, $g = 10 m s^{-2}$
    • Acceleration of the car, $a = + 10 m s^{-2}$ (downward)
    • (i) Speed $v = a t = 10 m s^{-2} \times 0.5 s = 5 m s^{-1}$
    • (ii) Average speed $= \frac{u + v}{2} = \frac{0 m s^{-1} + 5 m s^{-1}}{2} = 2.5 m s^{-1}$
    • (iii) Distance travelled, $s = \frac{1}{2} a t^2 = \frac{1}{2} \times 10 m s^{-2} \times (0.5 s)^2 = 1.25 m$
• Example 9.3: An object is thrown vertically upwards and rises to a height of 10 m. Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point.
  • Solution:
    • Distance travelled, $s = 10 m$
    • Final velocity, $v = 0 m s^{-1}$
    • Acceleration due to gravity, $g = 9.8 m s^{-2}$
    • Acceleration of the object, $a = -9.8 m s^{-2}$ (upward motion)
    • (i) $v^2 = u^2 + 2 a s$
      • $0 = u^2 + 2 \times (-9.8 m s^{-2}) \times 10 m$
      • $-u^2 = -2 \times 9.8 \times 10 m^2 s^{-2}$
      • $u = \sqrt{196} m s^{-1} = 14 m s^{-1}$
    • (ii) $v = u + a t$
      • $0 = 14 m s^{-1} - 9.8 m s^{-2} \times t$
      • $t = \frac{14}{9.8} s = 1.43 s$

Mass

• Mass is the measure of inertia.
• Greater the mass, the greater is the inertia.
• Mass remains the same whether the object is on the Earth, the Moon, or in outer space.
• The mass of an object is constant and does not change from place to place.

Weight

• The Earth attracts every object with a certain force.
• This force depends on the mass ($m$) of the object and the acceleration due to gravity ($g$).
• The weight of an object is the force with which it is attracted towards the Earth.
  • $F = m \times a$ $(9.13)$
  • $F = m \times g$ $(9.14)$
• The force of attraction of the Earth on an object is known as the weight of the object ($W$).
  • $W = m \times g$ $(9.15)$
• The SI unit of weight is the same as that of force: Newton (N).
• Weight is a force acting vertically downwards; it has both magnitude and direction.
• The value of $g$ is constant at a given place; therefore, at a given place, the weight of an object is directly proportional to its mass.
  • $W \propto m$
• The mass of an object remains the same everywhere, whereas its weight depends on location because $g$ depends on location.

Weight of an Object on the Moon

• The weight of an object on the Moon is the force with which the Moon attracts that object.
• The mass of the Moon is less than that of the Earth; hence, the Moon exerts lesser force of attraction on objects.
• Let the mass of an object be $m$. Let its weight on the Moon be $W<em>m$. Let the mass of the Moon be $M</em>m$ and its radius be $R_m$.
  • $W<em>m = G \frac{M</em>m \times m}{R_m^2}$ $(9.16)$
• Let the weight of the same object on the Earth be $W_e$. The mass of the Earth is $M$ and its radius is $R$.
  • $W_e = G \frac{M \times m}{R^2}$ $(9.17)$
• Substituting values from Table 9.1:
  • $W_m = G \frac{7.36 \times 10^{22} kg \times m}{(1.74 \times 10^6 m)^2} = 2.431 \times 10^{10} G m$ $(9.18a)$
  • $W_e = G \frac{5.98 \times 10^{24} kg \times m}{(6.37 \times 10^6 m)^2} = 1.474 \times 10^{11} G m$ $(9.18b)$
  • $\frac{W<em>m}{W</em>e} = \frac{2.431 \times 10^{10}}{1.474 \times 10^{11}} = 0.165$
    • $\frac{W<em>m}{W</em>e} = \frac{1}{6}$ $(9.19)$
• Weight of the object on the Moon = $\frac{1}{6}$ × its weight on the Earth.
• Example 9.4: Mass of an object is 10 kg. What is its weight on the Earth?
  • $m = 10 kg$
  • $g = 9.8 m s^{-2}$
  • $W = m \times g = 10 kg \times 9.8 m s^{-2} = 98 N$
• Example 9.5: An object weighs 10 N when measured on the surface of the Earth. What would be its weight when measured on the surface of the Moon?
  • $W<em>m = \frac{1}{6} \times W</em>e = \frac{1}{6} \times 10 N = 1.67 N$

Thrust and Pressure

• Thrust: Net force acting in a particular direction.
• Pressure: Force per unit area acting on the object.
• Situation 1: Fixing a poster on a bulletin board using drawing pins.
• Situation 2: Standing on loose sand versus lying down on the sand.
• The force acting on an object perpendicular to the surface is called thrust.
  • $Pressure = \frac{Thrust}{area}$ $(9.20)$
• The SI unit of pressure is $N/m^2$ or $N m^{-2}$, also called Pascal (Pa).
• Example 9.6: A block of wood (mass 5 kg, dimensions 40 cm × 20 cm × 10 cm) on a tabletop.
  • (a) 20 cm × 10 cm side:
    • Thrust = $F = m \times g = 5 kg \times 9.8 m s^{-2} = 49 N$
    • Area = $20 cm \times 10 cm = 200 cm^2 = 0.02 m^2$
    • Pressure = $\frac{49 N}{0.02 m^2} = 2450 N m^{-2}$
  • (b) 40 cm × 20 cm side:
    • Thrust = 49 N
    • Area = $40 cm \times 20 cm = 800 cm^2 = 0.08 m^2$
    • Pressure = $\frac{49 N}{0.08 m^2} = 612.5 N m^{-2}$
• The same force acting on a smaller area exerts a larger pressure, and a smaller pressure on a larger area.
• This is why nails have pointed tips, knives have sharp edges, and buildings have wide foundations.

Pressure in Fluids

• Liquids and gases are fluids.
• Fluids exert pressure on the base and walls of the container in which they are enclosed.
• Pressure exerted in any confined mass of fluid is transmitted undiminished in all directions.

Buoyancy

• Activity 9.4: Empty plastic bottle in a bucket of water.
• Water exerts a force on the bottle in the upward direction.
• The upward force exerted by the water goes on increasing as the bottle is pushed deeper until it is completely immersed.
• The upward force exerted by the water on the bottle is known as upthrust or buoyant force.
• All objects experience a force of buoyancy when they are immersed in a fluid.
• The magnitude of this buoyant force depends on the density of the fluid.

Why Objects Float or Sink When Placed on the Surface of Water?

• Activity 9.5: Iron nail and cork in a beaker filled with water.
• The nail sinks because the downward force acting on the nail is greater than the upthrust of water on the nail.
• Activity 9.6: Cork and iron nail of equal mass on the surface of water.
• The cork floats because the upthrust of water on the cork is greater than the weight of the cork.
• The density of a substance is defined as the mass per unit volume.
• Objects of density less than that of a liquid float on the liquid.
• Objects of density greater than that of a liquid sink in the liquid.

Archimedes’ Principle

• Activity 9.7: Stone tied to a rubber string or a spring balance, gradually lowered in water.
• The elongation of the string or the reading of the balance decreases as the stone is gradually lowered in the water.
• This indicates that some force acts on the stone in an upward direction: force of buoyancy.
• Archimedes' principle: When a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it.
• Archimedes used this principle to determine the purity of gold in the crown made for the king.
• Archimedes’ principle has many applications. It is used in designing ships and submarines.
• Lactometers and hydrometers are based on this principle.