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In this Physics tutorial, you will learn:

- What is gravitation?
- Where does gravitation differ from gravity, gravitational force and centre of gravity?
- What does the Newton's Law of Gravitation say?
- What are orbits?
- What makes objects rotate in orbits?
- How gravity affects our daily life?
- What are artificial satellites?
- How many types of artificial satellites are there? How do they differ from each other?

Have you ever thought why we stick on the ground but not to each other in normal conditions?

Why satellites do not fall on the ground?

Why the Moon stays always at the same distance from the Earth? Why these two celestial bodies do not collapse?

These questions and many others will get answer in this tutorial. It is an extension of our Physics tutorial "Types of Forces I. Gravitational Force and Weight." However, unlike the abovementioned tutorial, this tutorial deals only with celestial (macroscopic) bodies, not with normal ones we encounter in daily life.

As explained in tutorial "Types of Forces I. Gravitational Force and Weight", **gravity** is the attracting effect all objects possess, regardless their structure. Gravity produced by an object extends in all directions. The space around an object in which the gravity is observed is known as "gravitational field" and the force produced by this field is known as "gravitational force".

There is another related concept, which is often confused with gravity. It is known as **gravitation** and represents **a movement, or a tendency to move, towards a centre of gravity, as in the falling of bodies to the earth**.

Obviously, the concept of gravitation is related to that of **centre of gravity** (mass) we discussed in the tutorial "Centre of Mass and Types of Equilibrium". We explained it is a special point inside an object in which most properties of the object are concentrated.

In a few words, gravitation is the tendency to move towards the centre of gravity of an object due to the attracting effect of gravity produced by it. This means gravity is the cause, gravitation the effect, centre of gravity the target point, gravitational force the tool used to realize the objective and gravitational field the medium of event's occurrence.

In our tutorial "Types of Forces I. Gravitational Force and Weight", it was explained that Gravitational Force depends on the masses of the objects involved and the distance from their respective centres of gravity. The equation of gravitational force is

F_{g} = G × *m*_{1} × m_{2}*/**R*^{2}

where m_{1} and m^{2} are the masses of the objects involved, and R is the distance between the objects centres of gravity.

The quantity G is the **gravitational constant**. Its numerical value is

G = 6.674 × 10^{-11} *N × m*^{2}*/**kg*^{2}

The above equation is the mathematical expression of **Newton's Law of Gravitation**, which states that:

**Any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them**.

From this law and from the action-reaction principle, it is obvious that the interaction is mutual, i.e. gravitational force exerted by an object on another object, is equal in magnitude and opposite in direction to the gravitational force exerted by the second object into the first one.

What is the force by which the Sun attracts the Earth if the mass of the Sun is 1.989 × 1030 kg, mass of the Earth is 5.972 × 1024 kg and the distance from Sun to Earth is 151.95 million km.

First, we have to convert km to m and write the distance R in standard form. Thus,

R = 151.95 million km

= 151 950 000 km

= 151 950 000 000 m

= 1.5195 × 10^{11} m

= 151 950 000 km

= 151 950 000 000 m

= 1.5195 × 10

Now, let's calculate the gravitational force exerted by the Sun on the Earth using the equation of Newton's Law of Gravitation. Thus,

F_{g} = G × *M*_{Sun} × M_{Earth}*/**R*^{2}_{Sun-Earth}

=*6.674 × 10*_{-11} × 1.989 × 10^{30} × 5.972 × 10^{24}*/**(1.5195 × 10*^{11})^{2}

=*79.2758 × 10*^{43}*/**2.30888 × 10*^{22}

= 34.335 × 10^{21} N

= 3.4335 × 10^{22} N

=

=

= 34.335 × 10

= 3.4335 × 10

Since the attraction force between Sun and Earth is that big as seen in the above example, then why they don't collapse?

There are two factors, which prevent such a collapse. First, the presence of other celestial bodies around, produces other gravitational forces that tend to balance the gravitational force produced by the Sun. Second, the centripetal force produced when the Earth rotates around the Sun, makes the Earth rotate in a fixed circular path known as orbit.

Since centripetal and gravitational force have the same direction (both of them point towards the centre of curvature, it seem that they help each other. To fix this issue, we will use here the concept of centrifugal force, Fcf, which in essence is the same as centripetal force but in the opposite direction, i.e. it is a kind of reaction force. Thus, during a rotation in an orbit, the gravitational force is balanced by the centrifugal force. This helps us calculate the velocity of rotation necessary to keep an object (usually celestial bodies) moving in orbits.

Calculate the period of Earth revolution around the Sun if they were the only two celestial bodies in the universe. Take the known values: M_{S} = 1.989 × 1030 kg, M_{E} = 5.972 × 1024 kg and R_{S-E} = 151.95 million km.

Considering the Sun as stationary, it is obvious that equilibrium is ensured when gravitational force by which the Sun attracts the Earth towards itself is balanced by the centrifugal (the opposite of centripetal) force by which Earth tries to resist this approach to the Sun. These forces are equal in magnitude and opposite in direction, as shown in the figure below.

Let's write the clues first, by making the proper conversions in the units. Thus,

MS = 1.989 × 10^{30} kg

ME = 5.972 × 10^{24} kg

R_{S - E} = 151.95 million km = 1.5195 × 10^{11} m

G = 6.674 × 10^{-11} N × m2 / kg2

ME = 5.972 × 10

R

G = 6.674 × 10

Given that numerically we have

F_{g} = F_{cf}

Or

G × *M*_{Sun} × M_{Earth}*/**R*^{2}_{Sun-Earth} = *M*_{Earth} × v^{2}_{Earth}*/**R*_{Sun-Earth}

we obtain after simplifications and substitutions,

v_{Earth}^{2} = G × M_{Sun}/R_{Sun-Earth}

v_{Earth} = √**G × ***M*_{Sun}*/**R*_{Sun-Earth}

= √*6.674 × 10*^{-11} × 1.989 × 10^{3}0*/**1.5195 × 10*^{11}

= √**8.736 × 10**^{8}

= 2.9557 × 10^{4} *m**/**s*

= 29557 m/s

v

= √

= √

= 2.9557 × 10

= 29557 m/s

This is a constant value as all celestial bodies rotate uniformly around their own centre of rotation. Therefore, we must use the equation of uniform motion

L = v × t

where L is the path length and t is the moving time, to determine the period. Here we have

L = Perimeter of circle

= 2 × π × R_{Sun-Earth}

= 2 × 3.14 × 1.5195 × 10^{11} m

= 9.54246 × 10^{11} m

= 2 × π × R

= 2 × 3.14 × 1.5195 × 10

= 9.54246 × 10

Also, we insert the period of rotation T instead of time t in calculations because we are interested for one revolution only (see the Physics tutorial on Kinematics of Rotational Motion). Thus, we obtain

T = *2 × π × R*_{Sun-Earth}*/**v*

=*9.54246 × 10*^{11} m*/**2.9557 × 10*^{4}*m**/**s*

= 3.2284940 × 10^{7} s

= 32 284 940 s

=

= 3.2284940 × 10

= 32 284 940 s

Comparing this value with the true period or Earth revolution around the Sun (365.25 d × 24 h/d × 60 min/h × 60 s/min = 31 557 600 s), we observe that the difference is very small. Therefore, we draw as a conclusion that the other celestial bodies do not affect so much the process of Earth revolution around the Sun. This is because either they are very small compared to the Sun (the other planets or natural satellites of our solar system), or they are very far away from us (the other stars in the universe).

Since all objects attract each other because of gravitational force, then why we don't collapse with other people or objects around us?

The answer is simple. Gravitational force (as an attraction force between any two objects) depends on the object's dimensions and distance between them. Thus, if the gravitational force between Earth and a 50 kg object on its surface (using the simplified equation explained in the "Types of Forces I. Gravitational Force and Weight" tutorial, F_{g} = m × g), is

F_{g} = m × g

= 50 kg × 9.81*N**/**kg*

= 490.5 N

= 50 kg × 9.81

= 490.5 N

But when this object is 1 m away from another identical object on Earth surface, the gravitational force produced is

F_{g} = G × *m*_{1} × m_{2}*/**R*^{2}

=*6.674 × 10*^{-11} × 50 × 50*/**1*^{2}

= 1.6685 × 10^{-7} N

=

= 1.6685 × 10

or approximately 3 billion times smaller than the gravitational force exerted by the Earth. This is the reason why we neglect the effect of gravitational attraction between two common objects.

Artificial satellites are objects sent in the space by scientists for various purposes such as to observe natural phenomena such as weather, to distribute communication signals, for espionage, etc.

There are two types of artificial satellites: **geostationary** and **geosynchronous** satellites. Geo Synchronous satellites move in an orbit, which is synchronized with the earth's rotation to its own axis. This means that the orbital velocity of the satellite is equal to earth's rotational velocity. Any object placed in this orbit would return to the same point after one sidereal day. This implies, for any particular location in the earth (i.e.) fixed for a certain longitude, the satellite remain in a fixed area. However, for the orbit of geostationary satellites we can say it is a circular orbit placed above the equator at an altitude of 35786 km with zero inclination with horizontal plane. In this case, the satellite remains fixed at a particular point in the sky. Look at the figure.

Calculate the tangential velocity of a geosynchronous satellite if it is 3629 km above the equator. Take REarth = 6371 km.

In geosynchronous satellites, period of rotation is equal to that of the Earth, i.e.

T = 24 h × 60 *min**/**h* × 60 *s**/**min*

= 86400 s

= 86400 s

Radius of rotation is

r = R + h

= 6371 km + 3629 km

= 10000 km

= 10 000 000 m

= 6371 km + 3629 km

= 10000 km

= 10 000 000 m

Since the Earth (and therefore the satellite as well) rotate uniformly, we can use the equation of uniform motion to calculate the tangential velocity of satellite.

v = *2πR**/**T*

=*2 × 3.14 × 10000000 m**/**86400 s*

≈727 m/s

=

≈727 m/s

Obviously, the satellite must move faster than peripheral point of the Earth as the arc is wider. Its velocity at the given height is much greater than the tangential velocity of an object resting at equator (460 m/s). Therefore, such satellites need a powerful engine (and a high maintenance cost as a result) to operate regularly. This is the reason why not every country can send satellites in space.

**Gravitation** represents **a movement, or a tendency to move, towards a centre of gravity, as in the falling of bodies to the earth**.

In few words, gravitation is the tendency to move towards the centre of gravity of an object due to the attracting effect of gravity produced by it. This means gravity is the cause, gravitation the effect, centre of gravity the target point, gravitational force the tool used to realize the objective and gravitational field the medium of event's occurrence.

**Newton's Law of Gravitation**, states that:

**Any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them**.

Mathematically, the Newton's Law of Gravitation is written as:

F_{g} = G × *m*_{1} × m_{2}*/**R*^{2}

where m_{1} and m_{2} are the masses of the objects involved, and R is the distance between the objects centres of gravity.

The quantity G is the **gravitational constant**. Its numerical value is

G = 6.674 × 10^{-11} *N × m*^{2}*/**kg*^{2}

There are two factors, which prevent the celestial bodies of our solar system (and all the other objects in general) from collapsing due to the attraction effect of gravitational force. First, the presence of other celestial bodies around them produces other gravitational forces that tend to balance the gravitational force produced by the Sun. Second, the centripetal force produced when the Earth rotates around the Sun, makes the Earth (and other celestial bodies) rotate in a fixed circular path known as **orbit**.

During a rotation in an orbit, the gravitational force is balanced by the centrifugal force. This helps us calculate the velocity of rotation necessary to keep an object (usually celestial bodies) moving in orbits.

Gravitational force between two normal objects on earth is very small compared to that produced by Earth. As a result, objects do not collapse with each other but only with the Earth's surface.

Artificial satellites are objects sent in the space by scientists for various purposes such as to observe natural phenomena such as weather, to distribute communication signals, for espionage, etc.

There are two types of artificial satellites: **geostationary** and **geosynchronous** satellites. Geo Synchronous satellites move in an orbit, which is synchronized with the earth's rotation to its own axis, while for the orbit of geostationary satellites we can say it is a circular orbit placed above the equator at an altitude of 35786 km with zero inclination with horizontal plane. In this case, such satellites remain fixed at a particular point in the sky.

**1)** The orbital period of an artificial satellite revolving with engines turned off around the Moon at a distance of 1,737 km from the lunar centre is 2.0 hours. Calculate the mass of the Moon if no other factor but gravitational force between the Moon and satellite is taken into consideration.

- 5.974 × 10
^{18}kg - 3.985 × 10
^{22}kg - 5.974 × 10
^{22}kg - 3.985 × 10
^{18}kg

**Correct Answer: C**

**2)** At what tangential velocity must an object at 429 km above the Earth surface move to turn into an artificial satellite? Take radius of Earth 6371 km and mass of the Earth equal to 5.972 × 1024 kg.

- 6.802 × 10
^{3}m/s - 2.42 × 10
^{7}m/s - 2.42 × 10
^{3}m/s - 5.857 × 10
^{6}m/s

**Correct Answer: A**

**3)** Two satellites, one geostationary and the other geosynchronous are at the same position in a given instant. How far from each other will they be after 12 hours? Take the radius of Earth equal to 6371 km.

- 42157 km
- 84314 km
- 40010 km
- 91577 km

**Correct Answer: B**

We hope you found this Physics tutorial "Newton's Law of Gravitation" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In the next Physics section, we expand your insight and knowledge of "gravitation and its properties" with our Physics tutorials on Gravitational Potential Energy. Kepler Laws..

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