Understanding Gravity

Understanding Gravity

  1. When a boy accidentally drops a book from his hand. The book is pulled towards the Earth by the force of gravity.
    Understanding Gravity 1Fig. The book dropped by the boy is pulled towards the Earth
  2. The force of gravity on the Earth is caused by a gravitational field around the Earth. The gravitational field is a region in which an object experiences a force due to the gravitational attraction towards the centre of the Earth. Figure shows the direction of the gravitational field.Understanding Gravity 2Fig. The direction of the gravitational field around the Earth
  3. Gravitational acceleration is the acceleration of an object due to the pull of gravitational force.
  4. An object undergoes free fall if it is acted upon by gravitational force only. Near the surface of the Earth, all free falling objects will have a gravitational acceleration, g which is equal to 9.8 m s-2.
  5. Figure shows a stroboscopic photograph of a feather and an apple undergoing free fall in a vacuum. Both the feather and the apple fall with the same acceleration.
    (A stroboscopic photograph captures the image of a moving object and shows its respective positions at regular intervals of time.)
    Understanding Gravity 3
  6. The gravitational field strength at a point in the gravitational field is the gravitational force acting on a mass of 1 kg placed at that point. Therefore,
    Understanding Gravity 4
    Where, g = gravitational field strength,
    F = force of gravity, m = mass.
  7. The SI unit for gravitational field strength is N kg-1.

Also Read About: Newton’s Law of Universal Gravitation
Earth’s Gravitational Force

Weight

  1. The weight of an object is the force of gravity acting on the object.
  2. For an object with a mass, m, its weight, w is given by:
    w = mg
    where g is the gravitational field strength.
  3. The SI unit for weight is newton (N). Weight is a vector quantity.

Example 1. An astronaut has a mass of 75 kg. Calculate the gravitational force on the astronaut when he is standing
(a) on the Earth with a gravitational field strength of 9.8 N kg-1,
(b) on the Moon with a gravitational field strength of 1.6 N kg-1.
Solution:
Understanding Gravity 6

Example 2. The Hubble telescope has a mass of 11600 kg.
(a) Determine its weight when it is resting on Earth which has a gravitational field strength of 9.8 N kg-1.
(b) What is the value of the gravitational field strength if its weight at a particular orbit above the Earth is 95000 N?
Solution:
Understanding Gravity 9

Example 3. At a particular point above the Earth, an aircraft of mass 20 000 kg experiences a gravitational force of 192 000 N. Calculate the gravitational field strength at that point.
Solution: 

Understanding Gravity 5

Example 4. A ball is dropped from a building and undergoes free fall. What is its velocity just before it touches the ground which is 100 m from where it is dropped?
[g = 9.8 m s-2]
Solution:
Understanding Gravity 10

Activity 1

Aim: To determine the value of acceleration due to gravity.
Materials: Slotted masses, a piece of plank, cellophane tape, ticker tape
Apparatus: Ticker timer, 12 V a.c. power supply, stool, G-clamp
Method:
Understanding Gravity 7

  1. The apparatus is set up as shown in Figure.
  2. A strip of ticker tape about 2.5 m long is cut and passed through the ticker timer.
  3. One end of the tape is attached to the 100 g slotted mass.
  4. The ticker timer is switched on and the slotted mass is released so that it falls onto the plank.
  5. The tape is analysed to determine the gravitational acceleration, g.
  6. Steps 2 to 5 are repeated for slotted masses of mass, m = 200 g and 300 g.

Results:
Understanding Gravity 8
Discussion:

  1. The values of g for different masses are almost the same. This shows that the gravitational acceleration is constant. It does not depend on the value of the mass.
  2. The accurate value for g is 9.8 m s-2. The experimental values are smaller than this value due to unavoidable frictional forces and air resistance.

What Is Earth’s Gravitational Force

Earth’s Gravitational Force

  • The force which earth exerts on a body is called ‘force of gravity’. i.e.
    \(F=\frac{GMm}{{{R}^{2}}}\)
    Where M = mass of the earth, R = radius of the earth.
  • Due to this force, a body released from some height on the earth’s surface falls towards the earth with its velocity increasing at a constant rate.

Acceleration due to Gravity:

  • The acceleration produced in a body due to attraction of earth is called the acceleration due to gravity and is denoted by ‘g’.
  • \( g=\frac{GM}{{{R}^{2}}}=9.8\text{ m/}{{\text{s}}^{\text{2}}} \\ \)
    near the earth surface
  • \( \text{g on moon }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\approx \frac{{{g}_{e}}}{6}=\frac{9.8}{6}\text{m/}{{\text{s}}^{\text{2}}} \)
  • A body moving upwards with some initial velocity, experiences a retardation of 9.8 m/s2 & its velocity decreases continuously unless it becomes zero.
  • After this, it again starts falling towards the earth with the same acceleration of 9.8 m/s2.
  • The value of g is minimum at equator and maximum at poles.
  • The value of g does not depend upon the mass of the body falling towards the earth.

Variation In The Value Of Gravitational Acceleration (g)

(A) Variation with altitude or height:

  • When a body moves above the earth’s surface the distance of the body from the centre of earth increases there by decreasing the force of attraction.
  • At the earth’s surface
    \( \text{g}=\frac{\text{GM}}{{{\text{R}}^{\text{2}}}} \)
  • At a height h above the earth’s surface
    \( \text{g}=\frac{\text{GM}}{{{\text{(R + h)}}^{\text{2}}}} \)
    Variation-with-altitude-or-height
  • As we go above the earth’s surface the value of g goes on decreasing.

(B) Variation with depth d:

  • As we go deeper inside the earth, the body gets attracted by the core of the earth which is smaller in mass.
    Variation-with-depth-d
  • As we go inside the earth, the value of g decreases.
  • Force of attraction decreases and thus decreasing the value of g and becoming zero at the centre.

(C) Variation due to rotation of the earth:
Due to the rotation of the earth, the weight of a body is maximum at the poles and minimum at the equator.

Difference Between Mass and Weight

 

Mass

Weight

1.

Mass of a body is defined as the quantity of matter contained in it.Weight of a body is the force with which it is attracted towards the centre of the earth.
W = mg

2.

Mass of a body  remains constant and does not change from place to place.Weight of a body changes from place to place. It depends upon the value of g. Weight of a body on another planet will be different.

3.

Mass is measured by a pan balance.Weight is measured by a spring balance.

4.

Unit of mass is kg.Unit of weight is newton or kg-wt.

5.

Mass of a body cannot be zero.Weight of a body can be zero.

Ex. astronauts experience weightlessness in spaceships.

6.

Mass is a scalar quantity.Weight is a vector quantity.

Earth’s Gravitational Force Example Problems With Solutions

Example 1:    Given mass of earth is 6 × 1024 kg and mean radius of earth is 6.4 × 106 m. Calculate the value of acceleration due to gravity (g) on the surface of the earth.
Solution:    The formula for the acceleration due to gravity is given by
\( \text{g}=\frac{\text{GM}}{{{\text{R}}^{\text{2}}}} \)
Here, G = 6.67 × 10–11 Nm2/kg2;
M = mass of earth = 6 × 1024 kg;
R = radius of earth = 6.4 × 106 m
\(g=\frac{6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}}{{{(6.4\times {{10}^{6}})}^{2}}}\)
g = 9.8 m/s2

Example 2:    Calculate the value of acceleration due to gravity on a planet whose mass is 4 times as that of the earth and radius is 3 times as that of the earth.
Solution:    If M is the mass of the earth and R is the radius of earth, the value of acceleration due to gravity on the earth (ge) is given by
\( {{g}_{e}}=\frac{GM}{{{R}^{2}}}\text{ }…..\text{ (1)} \)
Let us consider a planet such that mass of the planet is equal to 4 times the mass of earth. Mp = 4M
Radius of the planet is equal to 3 times the radius of earth.
Re = 3R
Then, acceleration due to gravity on this planet(gp) is
\( {{g}_{e}}=\frac{G\times (4M)}{{{(3R)}^{2}}}=\frac{4}{9}.\frac{GM}{{{R}^{2}}}\text{ }…..\text{ (2)} \)
Dividing equation (2) by equation (1), we get
\( \frac{{{g}_{p}}}{{{g}_{e}}}=\frac{\frac{4}{9}\times \frac{GM}{{{R}^{2}}}}{\frac{GM}{{{R}^{2}}}} \)
\( \frac{{{g}_{p}}}{{{g}_{e}}}=\frac{4}{9} \)
\( {{g}_{p}}=\frac{4}{9}{{g}_{e}} \)
Since ge = 9.8 m/s2
\( {{g}_{p}}=\frac{4}{5}\times 9.8=4.35\text{ m/}{{\text{s}}^{\text{2}}} \)
Thus, acceleration due to gravity on the given planet is 4.35 m/s2.

Example 3:    Given the mass of the moon = 7.35 × 1022 kg and the radius of the moon = 1740 km. Calculate the acceleration experienced by a particle on the surface of the moon due to the gravitational force of the moon. Find the ratio of this acceleration to that experienced by the same particle on the surface of the earth.
Solution:    If Mm is the mass of the moon and Rm is its radius, then the acceleration experienced by a body on its surface is given by
\( a=\frac{G{{M}_{m}}}{R_{m}^{2}} \)
Here, Mm = 7.3 × 1022 kg;
Rm = 1740 km = 1.74 × 106 m
\( \therefore a=\frac{6.67\times {{10}^{-11}}\times 7.3\times {{10}^{22}}}{{{(1.74\times {{10}^{6}})}^{2}}}=1.57\text{ m/}{{\text{s}}^{\text{2}}} \)
While the acceleration due to gravity on the surface of the earth, is given by
\( {{g}_{e}}=\frac{G{{M}_{e}}}{R_{e}^{2}}=\frac{6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}}{{{(6.4\,\times \,{{10}^{6}})}^{2}}}=9.8\text{ }m/{{s}^{2}} \)
Comparing acceleration due to gravity on moon to that on the earth is
\( \frac{a}{g}=\frac{1.57}{9.8}=0.16 \)

Example 4:    At what height above the earth’s surface the value of g will be half of that on the earth’s surface ?
Solution:    We know that the value of g at earth’s surface is
\( g=\frac{GM}{{{R}^{2}}}\text{ }…..\text{ (1)} \)
While the value of g at a height h above the earth’s surface is given by
\( g\acute{\ }=\frac{GM}{{{(R+h)}^{2}}}\text{ }…..\text{ (2)} \)
Dividing equation (2) by equation (1), we get
\( \frac{g’}{g}={{\left( \frac{R}{R+h} \right)}^{2}}\text{ or }g\acute{\ }=g{{\left( \frac{R}{R+h} \right)}^{2}} \)
\( \text{Here},\text{ }g\acute{\ }=~\frac{g}{2} \)
\( \therefore \text{ }\frac{g}{2}=g{{\left( \frac{R}{R+h} \right)}^{2}} \)
\( \frac{R+h}{R}=\sqrt{2}R \)
or    R + h = √2 R
or    h = ( √2 – 1)R
or    h = (1.41 – 1) × 6400 = 0.41 × 6400
= 2624 km

Example 5:    Given mass of the planet Mars is 6 × 1023 kg and radius is 4.3 × 106 m. Calculate the weight of a man whose weight on earth is 600 N. (Given g on earth = 10 m/s2)
Solution:    Weight of the man on earth, W = mg
or 600 = m × 10 or m = 60 kg
So the mass of the man is 60 kg which will remain the same everywhere.
Now acceleration due to gravity on Mars,
\( {{g}_{m}}=\frac{G{{M}_{m}}}{R_{m}^{2}} \)
Here, Mm = 6 × 1023 kg; Rm = 4.3 × 106m
\( \therefore \text{ }{{g}_{m}}=\frac{6.67\times {{10}^{-11}}\times 6\times {{10}^{23}}}{{{(4.3\times {{10}^{6}})}^{2}}} \)
gm = 2.17 m/s2
Now, weight of the man on Mars will be
Wm = m × gm = 60 × 2.17 = 130.2 N