# Newton’s law of universal gravitation

In the subject of Newton’s law, was learned that every object which is initially rest becomes moves, or any object that initially moves becomes rest if there is “something” that moves or stops the object. Something is called “force”. Why does the fruit fall or move towards the surface of the earth after it is released from the stem? Newton’s law states that if the fruit moves, there must be a force acting on the fruit. The force that causes fruit or any object to fall towards the surface of the earth is called the force of gravity.

The problem of falling objects involving the force of gravity, which you are studying right now, has been thought of and studied by Isaac Newton, a British scientist. This problem that Newton thought had existed since ancient Greece. There are two basic problems that have been investigated by the Greeks, long before Newton was born. The question that is always questioned is why things always fall to the surface of the earth and how the planets move, including the sun and moon. The Greeks at that time saw the problem of falling objects and planetary movements as two different things. Thus it continued until Newton’s time. So what Newton produced was built on the work of people before him. What distinguishes Newton and the people before he is that Newton saw the problem of falling objects and planetary motion caused by one thing only and must obey the same law. In other words, Newton argued that the force that causes objects to fall to the surface of the ground is the same force, which causes the moon to move around the earth.

**The relationship between the force of gravity with distance and mass**

When formulating the law of gravity, Newton compared the acceleration of fruit falling near the surface of the earth with the centripetal acceleration of the moon when circling the earth. According to Newton, the acceleration experienced by falling fruit and the centripetal acceleration experienced by the moon circling the earth is caused by the same force, the earth’s gravitational force. Newton argues that the acceleration of the moon when it orbits the earth is proportional to 1/r2 where r = the distance between the center of the earth and the center of the moon = 3.84 x 10 8 m and the acceleration of falling fruit is proportional to 1/R^{2}, where R = distance between the center of the earth and fruit center. The fruit is near the surface of the earth so that R = R_{F}, where R_{E} = radius of earth = 6.37 x 10^{6} m. Thus, the comparison between moon acceleration (a_{M}) and fruit acceleration (g):

So the centripetal acceleration of the moon:

a_{M} = (2.75 x 10^{-4})(g) = (2.75 x 10^{-4})(9.8 m/s^{2}) = 2.7 x 10^{-3} m/s^{2}

Newton also calculated the centripetal acceleration of the moon in a different way. It is known that the lunar orbital period or the time month to do one revolution = 27.3 days = 2.36 x 10^{6} seconds. The length of the path traversed by the moon = the circumference of the lunar orbit = 2 (3.14) (r) where r = the distance between the center of the moon and the center of the earth. The centripetal acceleration of Moon:

The centripetal acceleration of the moon obtained in this calculation is almost the same as the results of previous calculations. Newton’s Second Law states that force is proportional to acceleration (F = m a). The above calculations show that acceleration is proportional to 1/r^{2} or acceleration is inversely proportional to the square of the distance (r^{2}). Thus it can be concluded that the gravitational force is proportional to 1/r^{2} or the gravitational force is inversely proportional to the square of the distance. Mathematically:

In addition to investigating the relationship between gravitational forces and the distance between objects, Newton also investigated the relationship between the gravitational force and the mass of objects. Newton’s Third Law states that if there is an action force there is a reaction force. The action force and reaction force have the same magnitude and opposite directions (F-action = -F reaction, a negative sign indicates that the direction of the force is opposite). If the earth’s gravitational force attracts the fruit, the gravitational force of the fruit also attracts the earth. Earth’s gravitational force works on the fruit, whereas the gravitational force of the fruit works on the earth. Likewise the gravitational force between the earth and the moon. Earth’s gravitational force with fruit’s gravitational force or gravitational force with the moon’s gravitational force is a reaction force. Because force is proportional to mass (F = m a) and because the force of action and the force of reaction have the same magnitude, the gravitational force between two objects must be proportional to the mass of the two objects. Mathematically:

**Newton’s law of universal gravitation**

Newton’s law of universal gravitation states that every object in the universe attracts one another, where the attraction between objects is proportional to their mass and inversely proportional to the square of their distance. If two objects with masses of m_{1} and m_{2} are separated by distance r then the magnitude of the gravitational force between the two objects is:

G = gravitational constant (6.673 x 10^{-11} Nm^{2}/ kg^{2}). G is measured by experiment.

F_{12} is the gravitational force applied to m_{1 }at m_{2} while F_{21} is the gravitational force applied to m_{2} at m_{1}. F_{12 }works at m_{2 }and pulls m_{2} towards m_{1}, while F_{21} works at m_{1} and pulls m_{1 }towards m_{2}. F_{21} and F_{12 }have the same magnitude and opposite direction. So F_{21} and F_{12} are pairs of action-reaction. r is the distance between the center of m_{1} and the center of m_{2}. The center of m_{1} is located in the center of the object, as is the center of m_{2}. If r is stated in km then it is first converted to meters (international system unit).

Sample problem 1 (gravitational force between 2 particles)

Calculate the gravitational force between two students, each with a mass of 30 kg and 40 kg and a distance of 1 meter.

Solution

The magnitude of the gravitational force is so small so that the two objects are not attracted to each other.

Sample problem 2 (gravitational force between 2 particles)

Calculate the total gravitational force experienced by m^{2 }due to the gravitational force applied to m1 at m^{2} and the gravitational force applied to m^{3} at m^{2}. The distance between center m^{1} and m^{2} is 2 meters and the distance between center m^{2} and m^{3} is 1 meter.

Solution:

The force of gravity of m_{1} at m_{2}:

The force of gravity of m_{3 }at m_{2}:

The total gravitational force experienced by m_{2}:

Fg total = (6.673 – 1.668) x 10^{-11} N = 5.005 x 10^{-11} N. In the figure below, the F_{32} arrow is longer than the F_{12 }arrow. So the direction of the total gravitational force goes to m_{3}.

Sample problem 3 (total gravity force = zero):

Two balls A and B, each with a mass of m and 5m. Both balls have the same diameter. If the gravitational force at a point between ball A and ball B is equal to zero, then the distance of that point from the surface of ball A is …

Pembahasan :

Solution:

The diameter of the ball A and B is 1 meter, so the radius of the ball A and B is 0.5 meters. The distance between the center of the ball A and B is 6 meters.

We place a test particle with mass m at a point located between ball A and ball B.

F_{A} = the gravitational force applied by ball A to the test particle (the direction of gravity towards the ball A).

F_{B} = the gravitational force that is applied by the ball B to the test particle (the direction of gravity towards the ball B).

So that the gravitational force experienced by the test particle is zero then F_{A }= F_{B }(F_{A} and F_{B} have opposite direction).

Use the quadratic formula to determine the value of x

So x = 4.88 meters

4.88 m – 0.5 m = 4.38 m. The gravitational force is zero, 4.38 meters from the surface of ball A. What is the distance from the surface of ball B?

**Measurement of universal gravity constant (G)**

To find out your body mass, you only need to stand on a scale and then read the scale of your body mass. It’s easy to measure your body mass. Well, how to measure the mass of the earth? There are no giant scales that are used to measure the mass of the earth and if there is no scale we cannot measure the mass of the earth using these scales because the earth is always moving at all times. The mass of the earth is known through calculation after the universal gravitational constant (G) is measured by experiment.

**Weight**

The weight of an object is the total gravitational force acting on an object due to the gravitational force applied by all objects in the universe on that object. If the object is located above the surface of the earth or the surface of the earth, then we ignore the gravitational force that is carried by other objects in the universe and consider the weight of an object as the gravitational force of the earth acting on the object.

w = weight of object, m_{E }= mass of earth, m = mass of object, R_{E} = radius of earth.

If an object is h above the surface of the earth (objects are above a certain height such as a plane, for example) or objects spaced r above the surface of the earth where r = R_{E} + h then the gravitational force acting on the object or the weight of the object is:

This equation shows that the weight of an object decreases to the square of the distance from the center of the earth. So the farther from the surface of the earth, the weight of objects decreases.

**Acceleration of gravity **

Based on Newton’s second law we know that the weight of an object is the force that causes the object to experience an acceleration of free fall (gravitational acceleration):

w = m g

Combine both equation w = F g and w = m g :

This equation explains the gravitational acceleration experienced by an object at a certain height above the earth’s surface. Through this equation, known that the acceleration of gravity decreases with increasing altitude and does not depend on the mass of objects falling freely.

Acceleration of gravity of objects located above the earth’s surface or above ground level:

The mass of the earth can be calculated by changing the above equation to:

Calculate the mass of the earth using this equation!

**Problems and Solutions**

1. Two objects m_{1} and m_{2} each with a mass of 6 kg and 9 kg separated by a distance of 5 cm. Object m_{3 }= 1 kg is placed between the two. If the gravitational force experienced by m_{3 }= 0 then the distance m_{3} from m1 is …

Gravitational force = 0 if F_{13} = F_{23}.

a = -1, b = -20, c = 50

Use quadratic formula:

The results cannot be negative. So x = r_{13} = distance m_{3} and m_{1} = 2.25 cm.

2. Two objects A and B are 60 cm apart. Mass A is 24 kg and mass B is 10 kg. Where is the location of a point that has a strong gravitational field equal to zero?

Gravitational field = 0 if g_{A} = g_{B}.

a = 7

b = -1440

c = 43200

Use quadratic formula:

x = r_{A} = 169.25 cm or

x = r_{A} = 36.46 cm