GRAVITATION CHAPTER # 06 PHYSICS 9TH – Short / Detailed Question Answers

 PHYSICS 9TH – Short / Detailed Question Answers

GRAVITATION
CHAPTER # 06
Q.1: State Newton’s Law of Gravitation. Derive the equation, $F = G\frac{m_1 m_2}{r^2}$.
Ans:
Newton’s Law of Gravitation:
Newton’s Law of universal gravitation states that: Every body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Derivation of Newton’s Law of Gravitation:
To understand this law, we consider two bodies of masses $m_1$ and $m_2$. The distance between their centres is $r$.

According to the statement, the force of attraction between two bodies is directly proportional to the product of their masses. Therefore:

$$F \propto m_1 m_2 \tag{(i)}$$

The gravitational force of attraction is inversely proportional to the square of the distance between the centres of the masses of the bodies. Therefore:

$$F \propto \frac{1}{r^2} \tag{(ii)}$$

Combining equation (i) and equation (ii):

$$F \propto \frac{m_1 m_2}{r^2}$$ $$F = G\frac{m_1 m_2}{r^2}$$

where $G$ is the constant of proportionality known as the “universal gravitational constant.” The value of $G$ in the SI unit is $6.673 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2}$. This is a very small value. ‘G’ remains constant everywhere.

We do not feel the gravitational force of attraction between objects around us due to the very small value of ‘G’. But it exists everywhere in the universe.

More Information:
Sir Isaac Newton was one of the greatest scientists in the world. He made fundamental contributions not only to the several branches of Physics (like optics and mechanics) but also to Astronomy and Mathematics. He formulated the laws of motion and the law of universal gravitation.

Q.2: What changes the path of the moon around the earth?
Ans: The pull of gravity changes the path of the Moon around the Earth.

Q.3: What are the characteristics / key points of gravitational force?
Ans:
Characteristics / Key Points of Gravitational Force:

Gravitational force has the following characteristics:

(i) It is always present between every two objects because of their masses.
(ii) It exists everywhere in the universe.
(iii) It forms an action–reaction pair.
(iv) It is independent of the medium between the objects.
(v) It is directly proportional to the product of the masses of objects.
(vi) It is inversely proportional to the square of the distance between the centres of the objects.
(vii) Hence, it follows the “Inverse Square Law”.

Q.4: Is Newton’s law of gravitation consistent with Newton’s third law of motion?
Ans:
Law of Gravitation and Newton’s Third Law of Motion:

According to Newton’s Law of gravitation, every two objects attract each other with equal force but in opposite directions. As shown in the given figure.

From the figure:

• $m_1$ = mass of body A
• $m_2$ = mass of body B
• $F_{12}$ = force with which body A attracts body B
• $F_{21}$ = force with which body B attracts body A

Then according to this law,
$F_{12} = -F_{21}$

This shows that the two forces are equal in magnitude but opposite in direction. Now, if $F_{12}$ is considered as “Action Force” and $F_{21}$ as “Reaction Force,” then by using the above equation, it is concluded that “Action equals reaction but in opposite direction.”

Recall that the above statement is in accordance with Newton’s third law of motion, which states that “To every action, there is always an equal and opposite reaction.”
Hence, Newton’s law of gravitation is consistent with Newton’s third law of motion.

For example: According to Newton’s law of universal gravitation, the Earth pulls the Moon with its gravity, and the Moon pulls the Earth with its gravity. Therefore, they form an action-reaction pair, which is in accordance with Newton’s third law of motion.

More Information:
Henry Cavendish in 1798 completed the 1st experiment that demonstrated Newton’s Law of universal gravitation. This happened more than a century after Newton had announced the law of universal gravitation.

Q.5: Define gravitational field. Describe the Earth’s gravitational field.
Ans:

Gravitational Field:
A gravitational field can be described as:
“A gravitational field is a region in which a mass experiences a force due to gravitational attraction.”

Earth’s Gravitational Field:
The Earth has an attractive gravitational field around it. Any object near the Earth experiences this force, which is due to Earth’s gravity. This field is directed towards the centre of the Earth.

This field is the strongest near the surface of the Earth and gets weaker as we move farther and farther away from the Earth. This force is called the "Field Force" because it acts on all objects whether they are in contact with Earth’s surface or not. So, it is a non-contact force. For example, it acts on an aeroplane whether it is standing on Earth’s surface or flying in the sky.

Gravitational Field Strength:
A body of mass one kilogram (1 kg) on Earth experiences a force of about ten Newton (10 N) due to Earth’s gravitational field. This force determines the gravitational field strength which is defined as Gravitational field strength ‘g’ is the gravitational force acting per unit mass.

The gravitational field strength ‘g’ is approximately 10 Newton per kilogram (10 Nkg$^-1$).
The gravitational field strength ‘g’ is different on different planets. For example, the gravitational field strength ‘g’ on the surface of the Moon is approximately 1.6 Newton per kilogram (1.6 Nkg$^-1$).
Acceleration due to gravity ‘g’ at different planets is shown in the following table.

Planet	Value of g (m/s$^2$)
Earth	10
Moon	1.62
Venus	8.87
Mars	3.77
Jupiter	25.95
Mercury	3.59
Saturn	11.08
Uranus	10.67
Neptune	14.07

Point of Ponder:
Does the whole solar system work in a push and pull network?
Gravity keeps things together. It is a force that attracts matter towards it. Anything with mass creates gravity, but the amount of gravity is proportional to the amount of mass. Therefore, Jupiter has a stronger gravitational pull than Mercury. Distance also affects the strength of the gravitational force. Therefore, the Earth has a stronger pull on us than Jupiter does, even though Jupiter is as big as over 1,300 Earths. While we are familiar with gravity’s impact on us and Earth, this force also has many effects on the entire solar system, too.

Creates Orbit:
One of the most noticeable effects of gravity in the solar system is the orbit of the planets. The sun could hold 1.3 million Earths, so its mass has a strong gravitational pull. When a planet tries to go past the sun at a high rate of speed, gravity grabs the planet and pulls it towards the sun. Likewise, the planet’s gravity is trying to pull the sun towards it but can’t because of the vast difference in mass. The planet keeps moving but is always caught up in the push-pull forces caused by the interaction of these gravitational forces. As a result, the planet begins orbiting the sun. The same phenomenon causes the moon to orbit around the Earth except for the Earth's gravitational force, not the sun’s, which keeps it moving around us.

Tidal Heating:
Just as the moon orbits the Earth, other planets have moons of their own. The push-pull relationship between the gravitational forces of the planets and their moons causes an effect known as tidal bulges. On Earth, we see these bulges as high and low tides because they occur over oceans. But on planets or moons without water, tidal bulges can occur over land. In some cases, the bulge created by gravity will be pulled back and forth because friction and waves in its distance from the primary source of gravity. The pulling causes friction and is known as tidal heating. On Io, one of Jupiter’s moons, tidal heating has caused volcanic activity. This heating may also be responsible for volcanic activity on Saturn’s Enceladus and liquid water underground on Jupiter’s Europa.

Creating Stars:
Giant molecular clouds made up of gas and dust slowly collapse because of the inward pull of their gravity. When these clouds collapse, they form lots of smaller areas of gas and dust that will eventually collapse as well. When these fragments collapse, they form stars. Because the fragments from the original GMC stay in the same general area, their collapse causes stars to form clusters.

Formation of Planets:
When a star is born, all of the dust and gas not needed in its formation ends up trapped in the orbit of the star. The dust particles have more mass than the gas, so they can begin to concentrate in certain areas where they come in contact with other dust grains. These grains are pulled together by their own gravitational forces and kept in orbit by the gravity of the star. As the collection of grains becomes bigger, other forces also begin to act upon it until a planet forms over a very long period of time.

Cause Destruction:
Because many things in the solar system are held together thanks to the gravitational pull among its components, strong external gravitational forces could pull those components apart, thus destroying the object. This happens with moons sometimes. For example, Neptune’s Triton is being pulled closer and closer to the planet as it orbits. When the moon gets too close, perhaps in 100 million to 1 billion years, the planet’s gravity will pull the moon apart. This effect might also explain the origin of the debris that makes up the rings found around all of the large planets, i.e., Jupiter, Saturn, and Uranus.

Q.6: What will be the effect on the gravitational pull between two objects if the medium between them is changed?
Ans:
The gravitational force present between the two masses will always remain the same no matter whether the masses are in air, vacuum, water, or separated by some hindrance. Because the gravitational force is not dependent on the intervening medium, there will be no effect on the gravitational pull between two objects if the medium between them is changed.

Q.7: Which force causes the moon to move in orbit around the earth?
Ans:
When an object moves in a circle at a constant speed, its direction constantly changes. A change in direction causes a change in velocity. This is because velocity is a vector quantity. It has an associated direction as well as a magnitude. A change in velocity results in acceleration, so an object moving in a circle is accelerating even though its speed may be constant.

An object will only accelerate if a resultant force acts on it. For an object moving in a circle, this resultant force is the centripetal force that acts towards the middle of the circle.

The gravitational attraction provides the centripetal force needed to keep planets in orbit around the Sun and all types of satellites in orbit around the Earth. The Earth’s gravity keeps the Moon orbiting us. It keeps changing the direction of the Moon’s velocity. This means gravity makes the Moon accelerate all the time, even though its speed remains constant.

Q.8: Define weight. What instrument is used to measure the weight of an object?
Ans:
We know that all the objects which are thrown upward in the air fall back to the ground. The force applied by the Earth’s gravitational field pulls the object downward. Weight is another name for Earth’s gravitational force on objects. Therefore, weight can be defined as:

The weight of an object is the measurement of the gravitational force acting on the object. Weight "W" of an object of mass ‘m’, in a gravitational field of strength ‘g’ is given by the relation:

$W = mg$

Like other forces, weight is also measured in Newton’s (N).

Instrument Used to Measure Weight:
A spring balance is used to measure the weight of an object. An object of mass 1 kg has a weight of 9.8 N near the surface of Earth. The objects with larger masses may have larger weights. Our weight varies slightly from place to place because Earth’s gravitational field strength varies at different places. The weight of the object changes as it moves away from the Earth. The weight of the object is different on different planets.

For example, we will have less weight on the Moon because the Moon’s gravitational field is weaker than Earth’s.

Q.9: What is the actual value of ‘g’ near the surface of Earth?
Ans:
Near Earth’s surface, gravitational acceleration is approximately $9.81 \, m/s^2$, which means that ignoring the effects of air resistance, the speed of an object falling freely will increase by about $9.81 \, m/s$ every second.

Q.10: If you go on a diet and lose weight, will you also lose mass? Explain.
Ans:
Let us imagine a surface that just barely surrounds our body as if we shrink—wrapped a body in plastic. By the law of conservation of mass, the only way our body can lose any amount of mass is for that amount of mass to pass out through the surface. So we just have to consider what bodily functions cause that to happen:

(i) Exhaling
(ii) Sweating
(iii) Excretion (In the non-technical sense of, roughly, things we do in the bathroom)

Any dead skin cells, strands of hair, etc., that fall off we would also count.
As a bonus, the “shrink-wrap view” also makes it easy to identify how we gain mass, by looking for all processes that cause matter to be drawn in through the invisible surface:

(i) Eating and drinking - solids and liquids through the oesophagus and gastrointestinal tract
(ii) Inhaling - gas through the trachea and lungs

The thing is, when most people talk about losing weight, they’re referring to a long-term average loss of mass, which means that the processes in the first list have to remove more mass over some extended period of time than the ones in the second list bring in. This requires some of the pre-existing mass in our body to be converted into the waste forms that we can dispose of through excretion, exhaling, and sweating. This pre-existing mass generally tends to be body fat. The other answers do a pretty good job filling in the details of how the fat gets converted to waste products.

Q.11: By using Newton’s Law of Gravitation, find the mass of the Earth.
Ans:
Mass of the Earth:
Mass of Earth cannot be measured directly by placing it on any weighing scale. But it can be measured by an indirect method. This method utilizes Newton’s Law of universal gravitation. Let us consider the following figure, in which a small ball is placed on the surface of the Earth.

m = Mass of the ball
Mₑ = Mass of the Earth
G = Universal gravitational constant
Rₑ = Radius of the Earth, which is also the distance between the ball and the center of the Earth.

Then according to Newton’s law of universal gravitation, the gravitational force F of the Earth acting on the ball is:

$F = G \frac{m Mₑ}{Rₑ^2}$ ...........(i)

whereas the force with which the Earth attracts the ball towards its center is equal to the weight of the ball. Therefore:

$F = W = mg$ ...........(ii)

Comparing equation (i) and (ii), we get:

$mg = G \frac{m Mₑ}{Rₑ^2}$

Re-arranged of above equations gives
$Mₑ = \frac{g Rₑ^2}{G}$ ..........(iii)

Numerical values of the constants at the right-hand side of equation (iii) are:
g = 10 Nkg⁻¹
Rₑ = 6.38 × 10⁶ m
G = 6.673 × 10⁻¹¹ Nm²kg⁻²

Substituting these values in equation (iii), we get:

$Mₑ = \frac{(10 Nkg⁻¹)(6.38 × 10⁶ m)²}{6.673 × 10⁻¹¹ Nm²kg⁻²} = 6.0 × 10²⁴ kg$

Thus, the mass of Earth is 6.0 × 10²⁴ kg.

Q.12: What are satellites? Define its types.
Ans:
Satellite:
A satellite is an object that revolves around a planet. Satellites are of two types:

(i) Natural satellites
(ii) Artificial satellites

Natural Satellites:
The planet which revolves around another planet naturally is called a “Natural Satellite”.

Example: Moon is a natural satellite because it revolves around the Earth naturally.

Artificial Satellites:
The objects which are sent into space by scientists to revolve around the Earth or other planets are called “Artificial Satellites”.

Examples: Sputnik–1 and Explorer–1 are amongst the artificial satellites.

Q.13: What are the uses of artificial satellites? OR Write down any four uses of an artificial satellite.
Ans:
Uses of Artificial Satellites:

Artificial satellites are used for different purposes like:

(i) for communication.
(ii) for making star maps.
(iii) for making maps of planetary surfaces.
(iv) for collecting weather information.
(v) for taking pictures of planets, etc.

Artificial satellites carry instruments, passengers, or both to perform different experiments in space. Rockets are used to put satellites into orbits in space.​

Q.14: What do you know about the orbits of artificial satellites?
Ans:
Orbits of Artificial Satellites:
Artificial satellites have been launched into different orbits around the Earth. There are different types of orbits like:

(i) For communication
(ii) Low–Earth orbit
(iii) Medium–Earth orbit
(iv) Geostationary orbit
(v) Elliptic orbit

These orbits are characterized based on different parameters like their distance from the Earth, their time period around the Earth, etc.

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Q.15: Describe communication satellites. OR What is a geostationary orbit?
Ans:
Communication Satellites:
An artificial satellite that completes its one revolution around the Earth in 24 hours is used for communication purposes. As Earth also completes its one rotation about its axis in 24 hours, therefore the above satellite appears to be stationary with respect to Earth. Its orbit is therefore called geostationary orbit. As it is used for communication purposes, it is known as a “Communication Satellite”. The height of a geostationary satellite is about 42300 km from the surface of the Earth. Its velocity with respect to Earth is zero.

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Interesting Information
Mass and radius of different objects:

Planet/Star	Mass (kg)	Radius (m)
Sun	1.99 × 10³⁰	6.96 × 10⁸
Moon	7.35 × 10²²	1.74 × 10⁶
Mercury	3.30 × 10²³	2.44 × 10⁶
Venus	4.87 × 10²⁴	6.05 × 10⁶
Earth	5.97 × 10²⁴	6.38 × 10⁶
Mars	6.42 × 10²³	3.40 × 10⁶
Jupiter	1.90 × 10²⁷	6.91 × 10⁷
Saturn	5.68 × 10²⁶	6.03 × 10⁷
Uranus	8.68 × 10²⁵	2.56 × 10⁷
Neptune	1.02 × 10²⁶	2.48 × 10⁷

Q.16: Define orbit.
Ans:
Orbit:
The curved path along which a natural or artificial satellite revolves around a planet is called an “orbit”.

Q.17: Derive an expression for the velocity that a satellite must possess when orbiting around the Earth and an expression for the time period of a satellite orbiting around the Earth.

Ans:
Expression for the velocity that a satellite must possess when orbiting around the Earth:

Newton’s law of gravitation has an important role in the motion of a satellite in its orbit because the gravitational pull of the Earth on the satellite provides the centripetal force needed to keep a satellite in orbit around some planet.

Let us consider the motion of a satellite that is revolving around the Earth. In the given figure:

• m = mass of the satellite
• M = mass of the Earth
• R = radius of the Earth
• r = R + h is radius of orbit
• h = height (altitude of satellite from the surface of the Earth)

We know that:
Centripetal force = Gravitational force

Or
$F_c = F_g$
$\frac{mv^2}{r} = \frac{GmM}{r^2}$

Substituting the value of $F_c$ and $F_g$ in equation (i):

$\frac{mv^2}{r} = \frac{GmM}{r^2}$
$v^2 = \frac{GM}{r}$
$v^2 = \frac{GM}{R + h}$

$v = \sqrt{\frac{GM}{R + h}}$
This gives the velocity that a satellite must possess when orbiting around Earth in an orbit of radius $r = R + h$.

This shows that the speed of the satellite is independent of its mass. Hence every satellite, whether it is very massive (large) or very light (small), has the same speed in the orbit.​

Expression for the time period of a satellite orbiting around the Earth:

The time required for a satellite to complete one revolution around the Earth in its orbit is called its time period “T”. The time period of a satellite can be calculated as:

The velocity of the satellite is given by equation (ii) as:

Substituting it in equation (iii):

The above equation gives the expression for the time period of a satellite orbiting around the Earth. Thus, Newton’s law of gravitation helps to describe the motion of a satellite in an orbit around the Earth.

Q.18: Describe the motion of an artificial satellite around the earth.

Ans:

Motion of Artificial Satellite around the Earth:

The satellites are put into their orbits around the Earth by rockets. When a satellite is put into orbit, its speed is selected carefully and correctly. If speed is not chosen correctly, then the satellite may fall back to Earth or its path may take it further into orbit. During the motion of a satellite in the orbit, the gravitational pull of the Earth on it is always directed towards the center of the Earth.

Newton used the following example to explain how gravity makes orbiting possible.

Let us imagine a cannonball launched from a high mountain. The given figure shows three paths the ball can follow:

Path A

• The cannonball is launched at a low speed.
• The cannonball will fall back to Earth.

Path B

• The cannonball is launched at a medium speed.
• The cannonball will fall back to Earth.

Path C

• The cannonball is launched at a high speed.
• The cannonball will not fall back to Earth. Instead, it orbits around the Earth.

The above example shows that for an artificial satellite to orbit the Earth and to retrace its path, it requires a certain orbital velocity.

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Q.19: Define orbital velocity. Derive an expression for the orbital velocity of an artificial satellite.

Ans:

Orbital Velocity:

The orbital velocity is defined as: The velocity required to keep the satellite into its orbit is called “Orbital Velocity”.
It is also defined as: The constant horizontal velocity is required to put the satellite into a stable circular orbit around the Earth.
It is also known as orbital speed or proper speed.

Expression for the orbital velocity of an artificial satellite:

The gravitational pull of Earth on a satellite provides the necessary centripetal force for orbital motion. Since this force is equal to the weight of the satellite:

$$W_s = mg, \text{ therefore }$$ $$F_c = W_s \,\,\, \text{(i)}$$

and $W_s = mgh$ where,

• $m =$ Mass of the satellite
• $g_h =$ Acceleration due to gravity at height 'h' from the surface of Earth.

The centripetal force $F_c$ on the satellite is:

$$F_c = \frac{mv^2}{r}$$

Substituting the values of $F_c$ and $W_s$ in equation (i):

$$\frac{mv^2}{r} = mgh$$

We know that:

If the satellite is orbiting very close to the surface of the Earth then:

$h \ll R$

In this case, orbital radius may be considered equal to the radius of Earth:

also $g_h = g$
and $v = v_c$

where,

• $v_c =$ critical velocity
• $g =$ acceleration due to gravity on the surface of the Earth

In terms of above factors, equation (ii) becomes:

This is known as "Critical Velocity".

If $g = 10 \, \text{ms}^{-2}$ and $R = 6.38 \times 10^6 \, \text{m}$

then equation (iii) becomes:

or:

or:

It should be noted that as the satellite gets closer to the Earth, the gravitational pull of the Earth on it gets stronger. So, the satellite to stay in an orbit closer to Earth needs to travel faster as compared to those satellites in the farther orbits.

Q.20: How would the value of $g$ and $G$ be affected, if the mass of the earth becomes four times?

Ans:

(i) As $g = \frac{GM_e}{R_e^2}$

It means that “$g$” is directly proportional to the mass of the Earth. If the mass of the Earth decreases, the value of “$g$” will also decrease. If the mass of the Earth increases, the value of “$g$” will also increase. If the mass of the Earth becomes four times, the value of “$g$” will also become four times. It means “$g$” will increase four times.

(ii) “$G$” is a gravitational constant; it remains the same throughout the universe.
If the mass of the Earth becomes four times, the value of “$G$” will not change. It will remain the same as it is a constant. Its value is $6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$.

Q.21: Define field force.

Ans: Field Force:

In physics, a force field is a vector field that describes a non-contact force acting on a particle at various positions in space.

Q.22: If you step on a scale and it gives a reading of 55 kg, is that a measure of your weight? If not, which physical quantity does it show?

Ans:

The terms “balance” and “scale” are often used interchangeably, and most of us would have a bit of trouble nailing down the characteristics of each.
There are technical and practical differences between the two, including exactly what is being measured and the types of applications they are used for. Although the terms “balance” and “scale” are used interchangeably, technically, they measure different things.

Balances Measure Mass:

Balances are instruments that measure mass (basically the amount of matter in something). A true balance measures mass directly by comparing the unknown mass to a known mass, a process that is not affected by changes in gravity. A balance of this sort will give the same reading irrespective of location because gravity will act on both sides of the balance equally.Scales Measure Weight:

Scales measure weight, which is the force acting on a mass that is equal to the object’s mass times its acceleration due to gravity. A scale can’t measure mass directly because the weighing mechanism and the weight of any given object are dependent on local gravity. On the Earth, gravitational acceleration can vary by as much as 0.5%, changing with distance from the Earth’s core, as well as the latitudinal way.

From a practical viewpoint, once a scale has been calibrated at its location, gravity is assumed or ignored. Therefore, weights are reported in mass units like kilograms, even though weight is a measurement of force using the newton. This also allows for the use of the terms “weight” to refer to both weight and mass, and “weigh” can be the process of determining either.

Q.23: If “$M_e$” is the mass of Earth, “$R_e$” radius of Earth, “$G$” is universal gravitational constant, then find the acceleration due to gravity “$g$”:

(i) on the surface of the Earth,
(ii) at the center of the Earth.

Ans:

(i) We consider a body of mass $m$ on the Earth’s surface. The gravitational force of attraction between the body and the Earth are equal to the weight of the body.

$$F = w = \frac{GM_e m}{R_e^2}$$ $$Mg = \frac{GM_e m}{R_e^2}$$ $$G = \frac{GM_e}{R_e^2}$$

We know that:
Mass of the Earth $M_e = 6.0 \times 10^{24} \, \text{kg}$
Radius of the Earth $R_e = 6.38 \times 10^{6} \, \text{m}$
Universal gravitational constant $G = 6.673 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$

Therefore,

$$g = \frac{GM_e}{R_e^2}$$ $$g = \frac{(6.673 \times 10^{-11})(6 \times 10^{24})}{(6.38 \times 10^{6})^2}$$ $$g = 9.8 \, \text{ms}^{-2}$$