Earth’s gravitational pull is often misunderstood, but without it, life on Earth would be impossible. In this lesson, we’ll define the gravitational pull and give some examples of how it is used. A quiz is provided to test your understanding.

## What Is Gravity?

Gravity – what is it? Is it a force? What causes it? Sir Isaac Newton was among the first to develop a model for gravity, purely through observation, but he could not explain it. Today, scientists still debate what causes gravity, and they are still hard-pressed to provide answers. We do know that if bodies did not have this gravitational attraction for one another, life as we know it on Earth would not exist.

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Here’s what the gravitational model tells us. Two bodies, anywhere in the universe, will experience an attraction for one another that is proportional to two things: 1) the product of their masses; and 2) the inverse square of the distance between their centers. Here, this attraction is represented by a mutual and equal pull, F, that exists between the two bodies: In this equation, G is the universal gravitational constant, each body has a known mass in kilograms, and r is the distance between their centers in meters.

## Definition of the Earth’s Gravitational Pull

The gravitational pull of the earth is the attraction that the earth exerts on an object or that an object exerts on the earth. It can be calculated using the mass of the object, the mass of Earth, the distance between the center of the object and the earth’s center, and the universal gravitational constant, a constant of proportionality that has been measured very accurately. It is equal to 6.674 x 10^-11.

One of the most common examples of this pull is the weight of an object on the surface of the earth. We will make the following assumptions:

• The mass of the earth is about 5.973 x 10^24 kg
• The green ball is sitting on the surface of the earth, at sea level and on the equator
• The equatorial radius of the earth is 6.378 x 10^6 m, which is the distance r

Here is the gravitational model equation with these values substituted: This famous result means that if we know the mass of an object on the surface of the earth, we will know how much pull the earth is exerting on it, that is, we will know its weight in Newtons (N). In order to get pounds, you divide Newtons by 4.448.

## Applications of the Earth’s Gravitational Pull

So what are the applications of knowing the earth’s gravitational pull?

Well first, we can determine your total body mass from your weight. Next time you’re on the scales, multiply your weight by 4.448 to get Newtons. Then divide by 9.8 and you will have your mass. Let’s say you weigh about 185 pounds. You would multiply 185 x 4.448, then divide by 9.8, which is about 84 kg. No matter what you may weigh elsewhere in the universe, your mass won’t change.

Next, we can determine the time of the fall of an object. When an object is dropped from a particular altitude, it accelerates downward based on the earth’s gravitational pull. In fact, the value of this acceleration is our old friend 9.8. If we neglect the friction of air, it is very easy to calculate the time of fall. In this expression, h is the starting height in meters: For example, if you drop an object from a height of 100 meters, it will take about 4.5 seconds to hit the ground.

For a final example, we can also determine satellite orbits.

Satellites are kept in orbit by the fact that the centrifugal force needed to keep them from leaving the orbit is equal to the gravitational pull. This is summarized very neatly by this equation, which scientists have used for decades: In this equation, v is the velocity the satellite needs to maintain as it orbits the earth and r is the orbital radius, in meters, from the satellite to the center of the earth. If you know v, then you can find r, or vice versa. A good example is the orbit of the GPS satellite constellation.

These 24 satellites must orbit the earth twice in a 24-hour period. They are positioned 12,552 miles above the surface of the earth, which gives them an orbital radius of r = 26,578,000 meters. From the designed equation, we get a velocity of v = 3,870 m/s. As long as this velocity is maintained, the satellites will remain in this orbit around the earth.

## Example of a Gravity Experiment

Let’s take a look at an example of how gravity works and is measured.

Let’s say Susie decides that she wants to create a project that simulates the earth’s gravitational pull on an object. She will do this by placing a large spherical object on a table and then suspending another smaller object nearby to see if there is an attraction between the two. The large object is a 16-lb bowling ball (mass = 7.26 kg) and the smaller object is a metal washer (mass = 0.02 kg). The radius of the bowling ball is 4.3 inches. The washer is dangled perfectly flat with respect to the surface of the bowling ball at a distance of 0.5 inches from the ball. Thus r is 4.8 inches or about 0.122 m. How much will the gravitational pull of the bowling ball on the washer be? Yikes! Susie finds that there will not be enough force to notice any effect at all. In fact, Susie’s family does not have enough money to purchase the kind of equipment needed to measure a force this small. What Susie did not realize is that gravitational pull is actually a very, very weak force!