Gravitation

Gravitational force holds artificial satellites in their orbit around the earth. They are launched from earth for purposes of scientific research, communication, weather forecasting, intelligence, etc. "Gravitation is a force of attraction that acts between all objects in the universe."

1.0Introduction

Gravitation is the weakest force in nature. It is negligible in the interactions of tiny particles, and thus plays no role in molecules, atoms, and nuclei. The gravitational attraction between objects of ordinary size, such as the gravitational force exerted by a building on a car, is too small to be noticed. When we consider very large objects, such as stars, planets, and satellites (moons), gravitation is of primary importance. The gravitational force exerted by the Earth on us and on the objects around us is a fundamental part of our experience. It is gravitation that binds us to the Earth and keeps the Earth and the other planets on course within the solar system. The gravitational force

2.0The Newtonian gravitation

Sir Isaac Newton did not discover gravitation, its effects have been known throughout human existence. But he was the first one to understand the broader significance of gravitation. Newton discovered that 'gravitation is universal, it is not restricted to Earth only', as others physicists of his time assumed. According to popular legend, Newton was sitting under an apple tree, when an apple landed on his head. He looked up at the sky, noticed the Moon, and reasoned that the same force that made the apple fall to the ground also kept the Moon in its orbit about the Earth.

The fact that the Moon did not move in a straight line suggested to him that there must be a force acting on it. Newton gave an idea that no-one else had even considered that 'gravity from the Earth acted through space and exerted a force on the Moon'.

3.0Newton's universal law of gravitation

'Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centers of the two objects'.

Let us consider two masses ${\mathrm{m}}_1$ and ${\mathrm{m}}_2$ lying at a separation distance r . Let the force of attraction between the two objects be F .

Newton's universal law of gravitation According to the universal law of gravitation, $\mathrm{F}\propto {\mathrm{m}}_1{\mathrm{m}}_2$ Also, $\mathrm{F}\propto \frac{1}{{\mathrm{r}}^2}$ (see figure) Combining both, we get, $F\propto \frac{m_1{m}_2}{r^2}$ or $\mathbf{F}=\frac{{\mathbf{Gm}}_1{\mathbf{m}}_2}{{\mathbf{r}}^2}$ where, G is the constant of proportionality and it is called the 'universal gravitational constant'.

4.0Definition of universal gravitational constant

Universal gravitation constant is the magnitude of the force (in newton) between a pair of 1 kg masses that are kept 1 meter apart.

5.0Unit of universal gravitational constant

According to the universal law of gravitation,

Thus, SI unit of $\mathrm{G}=\frac{\mathrm{Newton}\times (\text{ meter }{)}^2}{\mathrm{kg}\times \mathrm{kg}}=\frac{{\mathrm{Nm}}^2}{{\mathrm{kg}}^2}={\mathrm{Nm}}^2{\mathrm{kg}}^{-2}$ $\infty$ The numerical value of $G$ is $6.673\times 10^{-11}\mathrm{N}{\mathrm{m}}^2{\mathrm{kg}}^{-2}$.

6.0Characteristics of gravitational force

(1) It is a universal force of attraction. (2) It acts along the line joining the center of each mass. (3) It acts equally on each mass, i.e., it obeys Newton's third law. ${\mathrm{F}}_{12}=-{\mathrm{F}}_{21}$ (4) It is weaker if the masses are further apart. Gravitation acts in an inverse square manner, i.e $\mathrm{F}\propto \frac{1}{{\mathrm{r}}^2}$, where r is the distance between the centers of the masses. (5) It depends directly on the mass of each body involved, i.e. $\mathrm{F}\propto {\mathrm{m}}_1$ and $\mathrm{F}\propto {\mathrm{m}}_2$. (6) It is a long-range force i.e., its influence extends to very large distances. $\sigma$ Gravitational force between two masses does not depend on the medium present between them. Gravitational force between two masses does not depend on the presence of other masses around them.

• Gravitation is the weakest fundamental force that exist in nature.
• Q. Two objects $A$ and $B$ of masses $m_1$ and $m_2$ respectively are kept $r$ distance apart (given, ${\mathrm{m}}_1>{\mathrm{m}}_2$ ). (a) Which object will exert more gravitational force on the other? (b) If distance between them is made 6 r , how does the gravitational force between them change? (c) If an object $C$ of mass $m_3$ is placed between them, how does the gravitational force between object $A$ and $B$ change? Explanation: (a) Both will exert equal force on each other, they exist in an action-reaction pair. (b) If distance is made $6r$, the force between them becomes $\frac{1}{36}$ times the original value. This is because gravitational force follows inverse square relationship with the separation distance r. (c) Gravitational force between objects A and B remains the same as it does not depend on the other masses present near or in-between them.

7.0Importance of Newton's universal law of gravitation

Newton's universal law of gravitation successfully explained several phenomena which were believed to be unconnected: (1) The force that binds us to the Earth. (2) The motion of the Moon around the Earth. (3) The motion of planets around the Sun. (4) The tides due to the Moon and the Sun.

8.0Why Moon does not fall on Earth directly?

The motion of Moon is just like the motion of an object in circular motion. The velocity of Moon is directed tangential to the circle at every point along its path. The acceleration of Moon is directed towards the center of the circle i.e., towards the Earth (the central body) around which it is orbiting. This acceleration is caused by a centripetal force which is supplied by the gravitational force between the Earth and the Moon. If this force were absent, the Moon would continue in motion at the same speed in a direction tangential to the circular path and would have escaped away from the Earth. If the Moon had no tangential velocity, it would have fallen on Earth due to gravitation. Thus, it is the tangential velocity and the gravitational force that are perpendicular to each other, and cause the Moon to fall around the Earth without actually falling onto it (see figure).

Numerical Ability

To find the distance we should know the ${\mathrm{m}}_1,{\mathrm{m}}_2$ and F . Apply the formula $\mathrm{F}=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}$

• The mass of the Earth is $6\times 10^{24}\mathrm{kg}$ and that of the Moon is $7.4\times 10^{22}\mathrm{kg}$. If the distance between the Earth and the Moon is $3.84\times 10^5\mathrm{km}$, calculate the force exerted by the Earth on the Moon. ( $G=6.7\times 10^{-11}\mathrm{N}{\mathrm{m}}^2{\mathrm{kg}}^{-2}$ ) Solution: Mass of Earth, M $=6\times 10^{24}\mathrm{kg}$; mass of Moon, m $=7.4\times 10^{22}$ $\mathrm{kg},\mathrm{G}=6.7\times 10^{-11}\mathrm{N}{\mathrm{m}}^2{\mathrm{kg}}^{-2}$ Distance between Earth and Moon, r $=3.84\times 10^5\mathrm{km}=3.84\times$ $10^5\times 1000\mathrm{m}=3.84\times 10^8\mathrm{m}$ From universal law of gravitation, $\mathrm{F}=\frac{\mathrm{GMm}}{{\mathrm{r}}^2}=\frac{\left(6.7\times 10^{-11}\right)\left(6\times 10^{24}\right)\left(7.4\times 10^{22}\right)}{{\left(3.84\times 10^8\right)}^2}$ $=\frac{6.7\times 6\times 7.4\times 10^{24+22-11}}{3.84\times 3.84\times 10^{16}}=\frac{297.48\times 10^{35-16}}{14.746}$ $=20.1\times 10^{19}=2.01\times {10}^{20}\mathrm{N}$
• A planet of mass $M$ has two identical moons, each of mass $m$. Moon 1 is in a circular orbit of radius $r$ and Moon 2 is in a circular orbit of radius $2r$. What is the ratio of magnitude of the gravitational force exerted by the planet on Moon 1 to the magnitude of gravitational force exerted by the planet on Moon 2? Solution: Let the magnitudes of gravitational forces exerted by the planet on Moon 1 and Moon 2 are ${\mathrm{F}}_1$ and ${\mathrm{F}}_2$ respectively. Now, ${\mathrm{F}}_1=\frac{\mathrm{GMm}}{{\mathrm{r}}^2}$ and, ${\mathrm{F}}_2=\frac{\mathrm{GMm}}{(2\mathrm{r}{)}^2}$ or ${\mathrm{F}}_2=\frac{\mathrm{GMm}}{4{\mathrm{r}}^2}$ $\frac{(1)}{(2)}\Rightarrow \frac{F_1}{F_2}=\frac{\frac{GMm}{r^2}}{\frac{GMm}{4r^2}}=\frac{GMm}{r^2}\times \frac{4r^2}{GMm}=\frac{4}{1}$ $\therefore F_1:F_2=4:1$
• The gravitational force between two objects is 49 N . How much must the distance between these objects be decreased so that the force between them becomes double? Solution: Here, $\mathrm{F}=49\mathrm{N}$ Let $r=$ original distance between the objects, $m_1$ and ${\mathrm{m}}_2=$ masses of the objects Gravitational force between these objects is given by $\mathrm{F}=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}$ ...(1) Now, ${\mathrm{F}}^{\prime }=2\mathrm{F}=2\times 49\mathrm{N}=98\mathrm{N}$ Let, ${\mathrm{r}}^{\prime }=$ new distance between the objects, ${\mathrm{m}}_1$ and ${\mathrm{m}}_2=$ masses of the objects Gravitational force between these objects is given by ${\mathrm{F}}^{\prime }=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}$ Dividing (1) by (2), we get $\frac{\mathrm{F}}{{\mathrm{F}}^{\prime }}=\frac{{\mathrm{r}}^{\prime 2}}{{\mathrm{r}}^2}$ or $\frac{{\mathrm{r}}^{\prime }}{\mathrm{r}}=\sqrt{\frac{\mathrm{F}}{{\mathrm{F}}^{\prime }}}=\sqrt{\frac{\mathrm{F}}{2\mathrm{F}}}=\frac{1}{\sqrt{2}}\left[\square {\mathrm{F}}^{\prime }=2\mathrm{F}\right]$ $\therefore {\mathbf{r}}^{\prime }=\frac{\mathbf{r}}{\sqrt{2}}$

• Q. Which figure best represents the gravitational force acting on you and on Earth (Figure)? Explain.
  
  Explanation: The gravitational forces exist in action-reaction pair. The force on you should be towards the centre of Earth (or vertically downward). The force on earth will be opposite to that acting on you i.e., it should be away from the centre of Earth (vertically upwards). Thus, [figure (b)] best represents the gravitational force acting on you and on Earth.

9.0Kepler's laws of planetary motion

• Each planet in the Solar system moves in an elliptical orbit with the Sun at one focus (see figure).
  
• The line joining the planet and the Sun sweeps out equal areas in equal intervals of time (see figure).
  
  Kepler's Second law. For equal time intervals $\left(t_2-t_1\right)$ and $\left(t_4-t_3\right)$, Area A $=$ Area B.
• The square of time period of a planet revolving about the Sun is proportional to the cube of the mean distance of the planet from the Sun. In other words, 'the square of the orbital period of a planet is proportional to the cube of the semimajor axis of the elliptical orbit'. That is, ${\mathrm{T}}^2\propto {\mathrm{R}}^3$ or $\frac{{\mathbf{T}}^2}{{\mathbf{R}}^3}=$ Constant $\infty$ Let us consider two planets having orbital time periods ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$ respectively. Let their mean distances from the Sun are $R_1$ and $R_2$ respectively. Then, or $\frac{{\mathrm{T}}_1{}^2}{{\mathrm{T}}_2{}^2}=\frac{{\mathrm{R}}_1{}^3}{{\mathrm{R}}_2{}^3}$ or ${\left(\frac{{\mathbf{T}}_1}{{\mathbf{T}}_2}\right)}^2={\left(\frac{{\mathbf{R}}_1}{{\mathbf{R}}_2}\right)}^3$

• Q. According to Kepler's first law, planets follow elliptical orbits, with the Sun at one focus. How can you draw an ellipse? Explanation: To draw an ellipse, fix two pins on a piece of paper fixed on a cardboard or a drawing board. Two pins represent the two foci of the ellipse. Now, connect a length of string to the two pins. Use a pencil and the string to sketch out a smooth, closed curve (see figure). This closed curve is an ellipse.
  

10.0Free fall (motion under gravity)

Till 1600 AD, the teachings of the Greek philosopher Aristotle (384-322 BC) had held that heavier objects fall faster than lighter ones. The Italian physicist Galileo Galilei gave the present-day ideas of falling objects. Now, it is an established fact that, in the absence of air resistance, all objects dropped near the Earth's surface fall with the same constant acceleration under the influence of the Earth's gravity (see figure).

• Free fall is the motion of an object subject only to the influence of gravity. An object is in free fall as soon as it is dropped from rest, thrown downward or thrown upward.

Acceleration due to gravity

The constant acceleration of a freely falling body is called the acceleration due to gravity.

• The acceleration due to gravity is the acceleration of an object in free fall that results from the influence of Earth's gravity. Its magnitude is denoted with the letter ' g '.

Acceleration due to gravity at the surface of Earth

Let us consider an object of mass $m$ placed on the surface of Earth (see figure). Let the mass of Earth be M and radius of Earth be R. The gravitational force on the object due to Earth is given by, ${\mathrm{F}}_{\mathrm{g}}=\frac{\mathrm{GMm}}{{\mathrm{R}}^2}$

Let this force produces an acceleration ' $a$ ' on the object, then, ${\mathrm{Fg}}_{\mathrm{g}}=\mathrm{ma}$

From eq.(1) and eq.(2), we get, $\mathrm{ma}=\frac{\mathrm{GMm}}{{\mathrm{R}}^2}\text{ or }\mathrm{a}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}$

This acceleration is called acceleration due to gravity and it is denoted by g i.e., $\mathrm{a}=\mathrm{g}$ $\mathrm{g}=\frac{\mathbf{GM}}{{\mathbf{R}}^2}$

Calculating the value of g on Earth

Mass of the Earth $=6\times 10^{24}\mathrm{kg}$; radius of Earth $=6.4\times 10^6\mathrm{m};\mathrm{G}=6.673\times 10^{-11}\mathrm{N}{\mathrm{m}}^2{\mathrm{kg}}^{-2}$ Now, $g=\frac{\mathrm{GM}}{{\mathrm{R}}^2}=\frac{\left(6.67\times 10^{-11}\right)\left(6\times 10^{24}\right)}{{\left(6.4\times 10^6\right)}^2}=9.8{\mathbf{ms}}^{-2}$ $\infty$ The acceleration due to gravity g for any planet is (i) directly proportional to the mass of the planet (ii) inversely proportional to the square of the radius of the planet.

• Q. If the Earth were to suddenly shrink to half of its current radius with its mass remaining constant, would the acceleration due to gravity increase, decrease or stay the same? Explanation: Acceleration due to gravity ( g ) is directly proportional to the mass of Earth and inversely proportional to the square of its radius. Here, mass of Earth is given constant, while radius is shrunk to half of its current radius. Thus, acceleration due to gravity will increase and in this case it will be four times of its original value.

11.0Equations of motion of freely falling body

There are two main assumptions in free fall :

• Acceleration due to gravity ( g ) is constant throughout the motion and it acts vertically downwards.
• Air resistance is negligible.

Case 1

Case 2

Numerical Ability

• Q 1. A body falls freely from rest. Find (a) the distance it falls in 3 seconds (b) its speed after falling 20 m (c) the time required to reach a speed of $25\mathrm{m}/\mathrm{s}$ (d) time taken to fall $80\mathbf{m}$. Take $\mathbf{g}=10\mathrm{m}/{\mathrm{s}}^2$. Solution: Decode the problem Identify the equation $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$ (for part (i), since, $\mathrm{u},\mathrm{a}$ and t are given) $v^2=u^2+2$ as (for part (ii) since, $\mathrm{u},\mathrm{a}$ and s are given) $\mathrm{v}=\mathrm{u}+\mathrm{at}$ (for part (iii) since, $u$, $a$ and $v$ are given) $s=ut+\frac{1}{2}a{t}^2$ (for part (iv), since, $u,a$ and $s$ are given) Given, initial velocity, $u=0;\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$ (a) Distance travelled $\mathrm{s}=$ ? ; time $\mathrm{t}=3\mathrm{s}$

(b) Speed, $v=$ ? ; s = 20 m Now, $v^2=u^2+2\mathrm{gs}=(0{)}^2+2(10)(20)=400$ or $v=\sqrt{400}$ or $\mathbf{v}=20\mathrm{m}/\mathrm{s}$ (c) Time =? ; v=25 m/s Now, $v=u+gt$ or $25=0+(10)\mathrm{t}$ or $\mathbf{t}=2.5\mathbf{s}$ (d) Time $=$ ? ; distance travelled $\mathrm{s}=80\mathrm{m}$ Now, $s=ut+\frac{1}{2}{\mathrm{gt}}^2$ or $80=(0)t+\frac{1}{2}(10){\mathrm{t}}^2$ or $80=5{\mathrm{t}}^2$ or $t^2=16$ or $t=\sqrt{16}$ or $t=4\mathbf{s}$

• A boy drops a stone from the edge of the roof. It passes a window $2\mathbf{m}$ high in 0.1 s . How far is the roof above the top of the window? (Take $\mathrm{g}=10{\mathrm{ms}}^{-2}$ ) Solution:
  
  Decode the problem Identify the equation $s=ut+\frac{1}{2}a{t}^2$ (since, s , a and t are given) $v^2=u^2+2as$ (since, $u$, a and $v$ are given) Let a stone be dropped from the edge of the roof A. Let it passes over B with a velocity, say v. Consider motion BC. $\mathrm{u}=\mathrm{v},\mathrm{a}=10{\mathrm{ms}}^{-2};\mathrm{s}=2\mathrm{m};\mathrm{t}=0.1\mathrm{s}$ Using, $s=ut+\frac{1}{2}$ gt ${}^2$, we have $2=v(0.1)+\frac{1}{2}\times 10\times (0.1{)}^2$ $2=0.1\mathrm{v}+0.05\Rightarrow 0.1\mathrm{v}=2-0.05$ or $\mathrm{v}=\frac{1.95}{0.1}=19.5{\mathrm{ms}}^{-1}$ This initial velocity at $B$ in motion $BC$ is the final velocity in motion $AB$. Considering motion $AB$, we have $u=0;v=19.5{\mathrm{ms}}^{-1};\mathrm{s}=?;\mathrm{a}=10{\mathrm{ms}}^{-2}$ Using, ${\mathrm{v}}^2-{\mathrm{u}}^2=2$ as, we have $(19.5{)}^2-(0{)}^2=2\times 10\times s$ or $s=\frac{19.5\times 19.5}{2\times 10}$ or $\mathbf{s}=19.01\mathbf{m}\therefore$ Roof is 19.01 m above the window.
• A ball thrown up is caught by the thrower after 4 s . With what velocity was it thrown up? How high did it go? Where was it after $3s$ ? ( $\mathrm{g}=9.8{\mathbf{m}\mathbf{s}}^{-2}$ ) Solution: Decode the problem Identify the equation $\mathrm{v}=\mathrm{u}+\mathrm{at}$ (for part (i) since, t , a and v are given) ${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}$ (for part (ii) since, $u$, $a$ and $v$ are given) $s=ut+\frac{1}{2}a{t}^2$ (for part (iii), since, u , a and t are given) Since the time of going up is the same as that of coming down, therefore time of going up $=4/2=2$ s. Let it starts upward or $0=u-9.8\times 2$ or $\mathbf{u}=19.6{\mathbf{m}\mathbf{s}}^{-1}$ Again ${\mathrm{v}}^2-{\mathrm{u}}^2=2$ as $0-(19.6{)}^2=2(-9.8)h$ or $h=19.6\mathbf{m}$ After 2 s , it starts coming downwards (starting with $\mathrm{u}=0$ ). Considering downward motion,

$=4.9\mathrm{m}$ from top.

• Coconut is hanging on a tree at a height of 15 m from the ground. A boy launches a projectile vertically upwards with a velocity of $20\mathrm{m}{\mathrm{s}}^{-1}$. After what time the projectile will pass by the coconut? Explain the two answers in this problem. Solution: Decode the problem Identify the equation $s=ut+\frac{1}{2}a{t}^2$ (since, s , a and t are given) $v^2=u^2+2as$ (since, $u$, a and $v$ are given) Here $u=20\mathrm{m}{\mathrm{s}}^{-1};\mathrm{a}=-10\mathrm{m}{\mathrm{s}}^{-2};\mathrm{s}=15\mathrm{m};\mathrm{t}=$ ? Using $s=ut+\frac{1}{2}$ at ${}^2$, we have $15=20\mathrm{t}+(-10){\mathrm{t}}^2\Rightarrow 15=20\mathrm{t}-5{\mathrm{t}}^2$ or $3=4\mathrm{t}-{\mathrm{t}}^2$ or ${\mathrm{t}}^2-4\mathrm{t}+3=0$ or $(t-1)(t-3)=0$ $\because \mathrm{t}-1=0$ or $\mathbf{t}=1s$ $\mathrm{t}-3=0$ or $\mathbf{t}=3s$ After 1 s , it will cross coconut while going up and after 3 s , while coming down. While a ball is in free-fall, does is acceleration increase, decrease, or remain constant ? In free fall, acceleration due to gravity remains constant. It is velocity which decreases during upward motion and increases during downwards motion.

12.0Mass

The amount of matter contained in a body is called its mass.

• Mass of a body is constant at all places in the universe. It is measured by physical balance. It is a scalar quantity and it is always taken positive. Unit of mass SI unit : Kilogram (kg)CGS unit : Gram (g)

13.0Weight

The gravitational force exerted by the Earth on an object is called its weight. This force is directed towards the center of the Earth. It is denoted by letter W.

• The force with which an object is attracted towards the centre of Earth is called its weight. It is measured by a weighing machine or spring balance. Weight, $\mathrm{W}=\mathrm{mg}$ Where, $\mathrm{m}=$ mass of object ; $\mathrm{g}=$ acceleration due to gravity Unit of weight SI unit : Newton (N)CGS unit : Dyne

Variation of weight of an object

Since weight depends on g , weight varies with location. Objects weigh less at higher altitudes than at sea level, because $g$ decreases with increasing distance from the center of the Earth. Hence, weight, unlike mass, is not an inherent property of an object. Mass and weight are two different quantities. Mass of a body remains constant at all places, while weight of a body may change from place to place. Weight of a freely falling body is zero. At the centre of Earth, weight of a body is zero.

Weightlessness

We sense weight when something, such as the floor or chair, exerts a contact force on us. But if our chair and the floor, all are accelerating towards the Earth together like free fall, then no contact forces are exerted on us. Thus, our apparent weight is zero and we experience weightlessness. The space shuttle orbits nearly 400 km above Earth's surface. At this distance, $g=8.7\mathrm{m}/{\mathrm{s}}^2$, only slightly less than on Earth's surface. This means Earth's gravitational force is certainly not zero in the space shuttle. But, the astronauts experience weightlessness as the shuttle and everything in it falls freely towards the Earth. If a man is standing in a lift and the lift cable breaks, then the lift and the man will fall towards Earth with an acceleration g. The man is now in free fall, his apparent weight is zero and he will experience weightlessness.

• Q. (i) If you climb to the top of Mt. Everest, how does the acceleration due to gravity change as compared to that at sea level? (ii) Is there any change in your weight? (iii) Is their any change in your mass? Explanation: (i) As you climb to the top of Mt. Everest, distance (r) between you and the center of earth increases. As a result, the acceleration due to gravity ( g ) decreases, as it is inversely proportional to square of the distance (r) (see figure).
  
  (ii) Now, weight (W) of an object is directly proportional to acceleration due to gravity (g). Since, value of g decreases, your weight also decreases. (iii) Your mass will remain constant as it does not depend on g. Mass is simply the amount of matter contained in an object.

Weight of an object on Moon

We know that the weight of an object on the Earth is the force with which the earth attracts the object. Similarly, the weight of an object on the Moon is the force with which the Moon attracts that object. Let us consider an object of mass $m$. Let its weight on the Moon be $W_m$. Let the mass of the Moon be ${\mathrm{M}}_{\mathrm{m}}$ and its radius be ${\mathrm{R}}_{\mathrm{m}}$. By Newton's universal law of gravitation, the weight of the object on the Moon will be, ${\mathrm{W}}_{\mathrm{m}}=\frac{{\mathrm{GM}}_{\mathrm{m}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{m}}{}^2}$ Let weight of the same object on the Earth be $W_e$. Let the mass of the Earth be $M_e$ and its radius be Re. Thus, the weight of the object on the Earth will be, ${\mathrm{W}}_{\mathrm{e}}=\frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{e}}{}^2}$ $\frac{(1)}{(2)}\Rightarrow \frac{W_m}{W_e}=\frac{\frac{{\mathrm{GM}}_{\mathrm{m}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{m}}{}^2}}{\frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{e}}{}^2}}=\frac{{\mathrm{M}}_{\mathrm{m}}{R}_e{}^2}{{\mathrm{M}}_{\mathrm{e}}{\mathrm{R}}_{\mathrm{m}}{}^2}$ Now, mass of Earth is approximately 100 times the mass of Moon and radius of earth is nearly 4 times the radius of Moon i.e., $M_e=100M_m\text{; }$

Using (3) and (4), we get, $\frac{W_m}{W_e}=\frac{M_m{\left(4R_m\right)}^2}{100M_m{R}_m{}^2}=\frac{M_{m}16R_m{}^2}{100M_m{R}_m{}^2}=\frac{16}{100}\approx \frac{1}{6}$ or $\frac{{\mathbf{W}}_{\mathrm{m}}}{{\mathbf{W}}_{\mathrm{e}}}=\frac{1}{6}$ i.e., weight of the object on the Moon $=(1/6)\times$ its weight on the Earth

Now, ${\mathrm{W}}_{\mathrm{m}}={\mathrm{mg}}_{\mathrm{m}}$ and ${\mathrm{W}}_{\mathrm{e}}={\mathrm{mg}}_{\mathrm{e}}$, where ${\mathrm{g}}_{\mathrm{m}}$ and ${\mathrm{g}}_{\mathrm{e}}$ are accelerations due to gravity of Moon and Earth respectively. $\therefore \frac{{\mathrm{W}}_{\mathrm{m}}}{{\mathrm{W}}_{\mathrm{e}}}=\frac{{\mathrm{mg}}_{\mathrm{m}}}{{\mathrm{mg}}_{\mathrm{e}}}=\frac{1}{6}$ or $\frac{{\mathbf{g}}_m}{{\mathbf{g}}_e}=\frac{1}{6}$

Q. What is the weight of a person whose mass is $50\mathrm{kg}\left(\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2\right)$ ? Solution: Decode the problem Identify the equation $\mathrm{W}=\mathrm{mg}$ (since, $m$ and $g$ are given) The weight of the person $\mathrm{W}=\mathrm{mg}=50\times 9.8=490\mathrm{N}$

14.0Concept Map

15.0Some Basic Terms

• Artificial Satellites - An artificial satellite is an object that people have made and launched into orbit using rockets.
• Forecasting - Forecasting refers to the practice of predicting what will happen in the future by taking into consideration events in the past and present.
• Interactions - the action or influence of people, groups, or things on one another.
• Ordinary - normal; not unusual or different from others.
• Galaxies - A galaxy is a huge collection of gas, dust, and billions of stars and their solar systems.
• Significance - the importance or meaning of something.
• Restricted - controlled or limited.
• Physicists - Physicists explore the fundamental properties and laws that govern space, time, energy, and matter.
• Orbit - orbit is the curved trajectory of an object such as the trajectory of a planet around a star, or of a natural satellite around a planet, or of an artificial satellite around an object.
• Phenomena - a fact or an event in nature or society, especially one that is not fully understood.
• Tides - Tides are the rise and fall of sea levels caused by the combined effects of the gravitational forces exerted by the Moon (and to a much lesser extent, the Sun) and are also caused by the Earth and Moon orbiting one another.
• Influence - the power to affect, change or control somebody/something.
• Shrink - to become smaller or make something smaller.
• Physical balance - a physical balance is simply a weight measuring instrument. It is used to measure the weight of any unknown object with the help of the other object with known weight.
• Weighing machine - a measuring instrument for weighing; shows amount of mass.
• Spring balance - A spring balance is an instrument consisting of a hook attached to the end of a spring, used for weighing objects.
• Inherent - that is a basic or permanent part of somebody/something and that cannot be removed.
• Exerts - to make use of something, for example influence, strength, etc., to affect somebody/something.
• Apparent - that seems to be real or true but may not be.
• Space shuttle - a rocket-launched spacecraft able to land like an unpowered aircraft, used to make repeated journeys between the earth and space.