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Later in this chapter, we will see that the mass of other astronomical bodies also can be determined by the period of small satellites orbiting them. But until Cavendish determined the value of G , the masses of all these bodies were unknown.

Gravity above earth’s surface

What is the value of g 400 km above Earth’s surface, where the International Space Station is in orbit?

Strategy

Using the value of M E and noting the radius is r = R E + 400 km , we use [link] to find g .

From [link] we have

g = G M E r 2 = 6.67 × 10 −11 N · m 2 /kg 2 5.96 × 10 24 kg ( 6.37 × 10 6 + 400 × 10 3 m ) 2 = 8.67 m/s 2 .

Significance

We often see video of astronauts in space stations, apparently weightless. But clearly, the force of gravity is acting on them. Comparing the value of g we just calculated to that on Earth ( 9.80 m/s 2 ) , we see that the astronauts in the International Space Station still have 88% of their weight. They only appear to be weightless because they are in free fall. We will come back to this in Satellite Orbits and Energy .

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Check Your Understanding How does your weight at the top of a tall building compare with that on the first floor? Do you think engineers need to take into account the change in the value of g when designing structural support for a very tall building?

The tallest buildings in the world are all less than 1 km. Since g is proportional to the distance squared from Earth’s center, a simple ratio shows that the change in g at 1 km above Earth’s surface is less than 0.0001%. There would be no need to consider this in structural design.

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The gravitational field

[link] is a scalar equation, giving the magnitude of the gravitational acceleration as a function of the distance from the center of the mass that causes the acceleration. But we could have retained the vector form for the force of gravity in [link] , and written the acceleration in vector form as

g = G M r 2 r ^ .

We identify the vector field represented by g as the gravitational field    caused by mass M . We can picture the field as shown [link] . The lines are directed radially inward and are symmetrically distributed about the mass.

This figure shows a three dimensional vector graph. The x, y, z coordinate system is shown. A spherical mass M is shown at the origin and vectors are shown pointing toward it. The arrows decrease in length as their distance from the origin increases. A box, aligned with the coordinate axes, is also shown.
A three-dimensional representation of the gravitational field created by mass M . Note that the lines are uniformly distributed in all directions. (The box has been added only to aid in visualization.)

As is true for any vector field, the direction of g is parallel to the field lines at any point. The strength of g at any point is inversely proportional to the line spacing. Another way to state this is that the magnitude of the field in any region is proportional to the number of lines that pass through a unit surface area, effectively a density of lines. Since the lines are equally spaced in all directions, the number of lines per unit surface area at a distance r from the mass is the total number of lines divided by the surface area of a sphere of radius r , which is proportional to r 2 . Hence, this picture perfectly represents the inverse square law, in addition to indicating the direction of the field. In the field picture, we say that a mass m interacts with the gravitational field of mass M . We will use the concept of fields to great advantage in the later chapters on electromagnetism.

Practice Key Terms 2

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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