# Gravity Acceleration: What It Is, How To Measure It And Exercises

The acceleration of gravity  or gravitational acceleration is defined as the intensity of the gravitational field of the Earth. That is, the force it exerts on any object, per unit mass.

It is denoted by the now familiar letter g and its approximate value near the earth’s surface is 9.8 m / s 2 . This value may undergo small variations with geographic latitude and also with height with respect to sea level.

The acceleration of gravity, in addition to having the aforementioned magnitude, has direction and sense. In effect, it is directed vertically towards the center of the earth.

The gravitational field of the Earth can be represented as a set of radial lines that point towards the center, as shown in the previous figure.

Article index

• one

What is the acceleration of gravity?

• 1.1

The Law of Universal Gravitation

• two

How is gravity measured on different planets?

• 2.1

Experiment to determine the value of g

• 2.2

Standard value of g on Earth, on the Moon and on Mars

• 23

Gravity on the moon

• 2.4

Gravity on mars

• 3

Exercise solved: the falling apple

• 4

References

## What is the acceleration of gravity?

The value of the acceleration of gravity on Earth or on any other planet is equivalent to the intensity of the gravitational field it produces, which does not depend on the objects around it, but only on its own mass and radius.

The acceleration of gravity is often defined as the acceleration experienced by any object in free fall in the vicinity of the earth’s surface.

In practice, this is what almost always happens, as we will see in the following sections, in which Newton’s Law of Universal Gravitation will be used.

Newton is said to have discovered this famous law while meditating on falling bodies under a tree. When he felt the hit of the apple on his head, he immediately knew that the force that makes the apple fall is the same that causes the Moon to orbit the Earth.

### The Law of Universal Gravitation

Whether the legend of the apple was true or not, Newton realized that the magnitude of the gravitational force of attraction between any two objects, for example between the Earth and the Moon, or the Earth and the apple, must depend on their masses. :

$F\propto&space;m_{1}.m_{2}$

$F\propto&space;\frac{1}{r^{2}}$

$F\propto&space;\frac{m_{1}.m_{2}}{r^{2}}$

$F=G&space;\frac{m_{1}.m_{2}}{r^{2}}$

#### Characteristics of the gravitational force

The gravitational force is always attractive; that is, the two bodies it affects attract each other. The opposite is not possible, since the orbits of the celestial bodies are closed or open (comets, for example) and a repulsive force can never produce a closed orbit. So the masses always attract each other, whatever happens.

A fairly good approximation to the actual shape of the Earth (m 1 ) and of the Moon or apple (m 2 ) is to assume that they are spherical in shape. The following figure is a representation of this phenomenon.

Here both the force exerted by m 1 on m 2 and the force exerted by m 2 on m 1 are represented , both of equal magnitude and directed along the line joining the centers. They are not canceled, since they are applied to different objects.

In all the following sections it is assumed that the objects are homogeneous and spherical, therefore their center of gravity coincides with their geometric center. The entire mass can be assumed to be concentrated right there.

## How is gravity measured on different planets?

Gravity can be measured with a gravimeter, a gravity measurement device used in geophysical gravimetric surveys. Currently they are much more sophisticated than the originals, but at the beginning they were based on the pendulum.

The pendulum consists of a thin, light and inextensible rope of length L. One of its ends is fixed to a support and a mass m is hung from the other.

When the system is in equilibrium, the mass hangs vertically, but when it is separated from it, it begins to oscillate, executing a back and forth movement. Gravity is responsible for it. For all that follows, it is valid to assume that gravity is the only force acting on the pendulum.

The period T of oscillation of the pendulum for small oscillations is given by the following equation:

$T=2\pi&space;\sqrt{\frac{L}{g}}$

$T^{2}=4\pi&space;^{2}\frac{L}{g}=\left&space;(\frac{4\pi&space;^{2}}{g}&space;\right&space;)L$

### Experiment to determine the value of g

#### Materials

– 1 metal ball.

– Rope of several different lengths, at least 5.

– Measuring tape.

– Transporter.

– Stopwatch.

– A support to fix the pendulum.

– Graph paper or computer program with spreadsheet.

#### Process

1. Select one of the strings and assemble the pendulum. Measure the length of the string + the radius of the sphere. This will be the length L.
2. Remove the pendulum from the equilibrium position about 5 degrees (measure it with the protractor) and let it swing.
3. Simultaneously start the stopwatch and measure the time of 10 oscillations. Write down the result.
4. Repeat the above procedure for the other lengths.
5. Find the time T it takes for the pendulum to swing (dividing each of the above results by 10).
6. Square each value obtained, obtaining T 2
7. On graph paper, plot each value of T 2 on the vertical axis, against the respective value of L on the horizontal axis. Be consistent with the units and do not forget to take into account the error of appreciation of the instruments used: tape measure and stopwatch.
8. Draw the best line that fits the plotted points.
9. Find the slope m of said line using two points that belong to it (not necessarily experimental points). Add the experimental error.
10. The above steps can be accomplished with a spreadsheet and the option to construct and fit a straight line.
11. From the value of the slope to clear the value of g with its respective experimental uncertainty.

### Standard value of g on Earth, on the Moon and on Mars

The standard value of gravity on Earth is: 9.81 m / s 2 , at 45º north latitude and at sea level. Since the Earth is not a perfect sphere, the values ​​of g vary slightly, being higher at the poles and lower at the equator.

Those who want to know the value in their locality can find it updated on the website of the German Metrology Institute PTB ( Physikalisch-Technische Bundesanstalt ), in the Gravity Information System (GIS) section.

### Gravity on the moon

The gravitational field of the Moon has been determined by analyzing radio signals from space probes orbiting the satellite. Its value on the lunar surface is 1.62 m / s 2

### Gravity on mars

The value of g P for a planet depends on its mass M and its radius R as follows:

$mg_{P}=G\frac{mM}{R^{2}}$

Therefore:

$g_{P}=G\frac{M}{R^{2}}$

For the planet Mars, the following data is available:

M = 6.4185 x 10 23 kg

R = 3390 km

G = 6.67 x 10 -11 Nm 2 / kg 2

With these data, we know that the gravity of Mars is 3.71 m / s 2 . Naturally the same equation can be applied with the data of the Moon or any other planet and thus estimate the value of its gravity.

## Exercise solved: the falling apple

Suppose that both the Earth and an apple are spherical in shape. The mass of the Earth is M = 5.98 x 10 24 kg and its radius is R = 6.37 x 10 6  m. The mass of the apple is m = 0.10 kg. Suppose there is no other force except gravity. From Newton’s Law of Universal Gravitation find:

a) The gravitational force that the Earth exerts on the apple.

b) The acceleration experienced by the apple when it is released from a certain height, according to Newton’s Second Law.

#### Solution

a) The apple (supposedly spherical, like the Earth) has a very small radius compared to the Earth’s radius and is immersed in its gravitational field. The following figure is obviously not to scale, but there is a diagram of the gravitational field g , and the force F exerted by the earth on the apple:

By applying Newton’s Law of Universal Gravitation, the distance between the centers can be considered roughly the same value as the Earth’s radius (the height from which the apple falls is also negligible compared to the Earth’s radius). Therefore:

b) According to Newton’s Second Law, the magnitude of the force exerted on the apple is:

F = ma = mg

Whose value is 0.983 N, according to the previous calculation. Equating both values ​​and then solving for the magnitude of the acceleration, we obtain:

mg = 0.983 N

g = 0.983 N / 0.10 kg = 9.83 m / s 2

This is a very good approximation to the standard value of gravity.

## References

1. Giancoli, D. (2006). Physics: Principles with applications. Sixth Edition. Prentice Hall. 118-122.
2. Hewitt, Paul. (2012). Conceptual Physical Science. Fifth Edition. Pearson. 91-94.
3. Rex, A. (2011). Fundamentals of Physics. Pearson. 213-221.