Physics Tutorial: Newton's Law of Gravitation

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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?
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8.1Newton's Law of Gravitation

Introduction

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.

Gravity, Centre of Gravity and Gravitation

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.

Recap on Gravitational Force. Newton's Law of Gravitation

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

Fg = G × m1 × m2/R2

where m1 and m2 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 × m2/kg2

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.

Example 1

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.

Solution 1

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 × 1011 m

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

Fg = G × MSun × MEarth/R2Sun-Earth
= 6.674 × 10-11 × 1.989 × 1030 × 5.972 × 1024/(1.5195 × 1011)2
= 79.2758 × 1043/2.30888 × 1022
= 34.335 × 1021 N
= 3.4335 × 1022 N

What Makes Objects Rotating in Orbits?

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.

Example 2

Calculate the period of Earth revolution around the Sun if they were the only two celestial bodies in the universe. Take the known values: MS = 1.989 × 1030 kg, ME = 5.972 × 1024 kg and RS-E = 151.95 million km. Physics Tutorials: This image provides visual information for the physics tutorial Newton's Law of Gravitation

Solution 2

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.

Physics Tutorials: This image provides visual information for the physics tutorial Newton's Law of Gravitation

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

MS = 1.989 × 1030 kg
ME = 5.972 × 1024 kg
RS - E = 151.95 million km = 1.5195 × 1011 m
G = 6.674 × 10-11 N × m2 / kg2

Given that numerically we have

Fg = Fcf

Or

G × MSun × MEarth/R2Sun-Earth = MEarth × v2Earth/RSun-Earth

we obtain after simplifications and substitutions,

vEarth2 = G × MSun/RSun-Earth
vEarth = √G × MSun/RSun-Earth
= √6.674 × 10-11 × 1.989 × 1030/1.5195 × 1011
= √8.736 × 108
= 2.9557 × 104 m/s
= 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 × RSun-Earth
= 2 × 3.14 × 1.5195 × 1011 m
= 9.54246 × 1011 m

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 × RSun-Earth/v
= 9.54246 × 1011 m/2.9557 × 104m/s
= 3.2284940 × 107 s
= 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).

Gravity in Daily Life

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, Fg = m × g), is

Fg = m × g
= 50 kg × 9.81 N/kg
= 490.5 N

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

Fg = G × m1 × m2/R2
= 6.674 × 10-11 × 50 × 50/12
= 1.6685 × 10-7 N

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.

The Two Types of Artificial Satellites

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.

Physics Tutorials: This image provides visual information for the physics tutorial Newton's Law of Gravitation Physics Tutorials: This image provides visual information for the physics tutorial Newton's Law of Gravitation

Example 3

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

Physics Tutorials: This image provides visual information for the physics tutorial Newton's Law of Gravitation

Solution 3

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

Radius of rotation is

r = R + h
= 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

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.

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