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By the end of this section, you will be able to:
  • Explain the connection between the constants G and g
  • Determine the mass of an astronomical body from free-fall acceleration at its surface
  • Describe how the value of g varies due to location and Earth’s rotation

In this section, we observe how Newton’s law of gravitation applies at the surface of a planet and how it connects with what we learned earlier about free fall. We also examine the gravitational effects within spherical bodies.

Weight

Recall that the acceleration of a free-falling object near Earth’s surface is approximately g = 9.80 m/s 2 . The force causing this acceleration is called the weight of the object, and from Newton’s second law, it has the value mg . This weight is present regardless of whether the object is in free fall. We now know that this force is the gravitational force between the object and Earth. If we substitute mg for the magnitude of F 12 in Newton’s law of universal gravitation, m for m 1 , and M E for m 2 , we obtain the scalar equation

m g = G m M E r 2

where r is the distance between the centers of mass of the object and Earth. The average radius of Earth is about 6370 km. Hence, for objects within a few kilometers of Earth’s surface, we can take r = R E ( [link] ). The mass m of the object cancels, leaving

g = G M E r 2 .

This explains why all masses free fall with the same acceleration. We have ignored the fact that Earth also accelerates toward the falling object, but that is acceptable as long as the mass of Earth is much larger than that of the object.

This figure shows an illustration of the earth, with a building on its surface. A cut away of a quarter of the earth shows several layers. The center of the earth is labeled C M, and the radius from the center to the building is labeled R E. An enlarged view of the building and a portion of the earth is also shown. In this view, we see that the arrow labeled R E terminates in the building, slightly above the surface of the earth.
We can take the distance between the centers of mass of Earth and an object on its surface to be the radius of Earth, provided that its size is much less than the radius of Earth.

Masses of earth and moon

Have you ever wondered how we know the mass of Earth? We certainly can’t place it on a scale. The values of g and the radius of Earth were measured with reasonable accuracy centuries ago.

  1. Use the standard values of g , R E , and [link] to find the mass of Earth.
  2. Estimate the value of g on the Moon. Use the fact that the Moon has a radius of about 1700 km (a value of this accuracy was determined many centuries ago) and assume it has the same average density as Earth, 5500 kg/m 3 .

Strategy

With the known values of g and R E , we can use [link] to find M E . For the Moon, we use the assumption of equal average density to determine the mass from a ratio of the volumes of Earth and the Moon.

Solution

  1. Rearranging [link] , we have
    M E = g R E 2 G = 9.80 m/s 2 ( 6.37 × 10 6 m ) 2 6.67 × 10 −11 N · m 2 /kg 2 = 5.95 × 10 24 kg.
  2. The volume of a sphere is proportional to the radius cubed, so a simple ratio gives us
    M M M E = R M 3 R E 3 M M = ( ( 1.7 × 10 6 m ) 3 ( 6.37 × 10 6 m ) 3 ) ( 5.95 × 10 24 kg ) = 1.1 × 10 23 kg.

    We now use [link] .
    g M = G M M r M 2 = ( 6.67 × 10 −11 N · m 2 /kg 2 ) ( 1.1 × 10 23 kg ) ( 1.7 × 10 6 m ) 2 = 2.5 m/s 2

Significance

As soon as Cavendish determined the value of G in 1798, the mass of Earth could be calculated. (In fact, that was the ultimate purpose of Cavendish’s experiment in the first place.) The value we calculated for g of the Moon is incorrect. The average density of the Moon is actually only 3340 kg/m 3 and g = 1.6 m/s 2 at the surface. Newton attempted to measure the mass of the Moon by comparing the effect of the Sun on Earth’s ocean tides compared to that of the Moon. His value was a factor of two too small. The most accurate values for g and the mass of the Moon come from tracking the motion of spacecraft that have orbited the Moon. But the mass of the Moon can actually be determined accurately without going to the Moon. Earth and the Moon orbit about a common center of mass, and careful astronomical measurements can determine that location. The ratio of the Moon’s mass to Earth’s is the ratio of [the distance from the common center of mass to the Moon’s center] to [the distance from the common center of mass to Earth’s center].

Questions & Answers

if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand
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hey , can you please explain oxidation reaction and redox ?
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for grade 12 or grade 11?
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it is the force or component of the force that the surface exert on an object incontact with it and which acts perpendicular to the surface
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a rate of change in velocity of an object whith respect to time
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Acceleration is a rate of change in velocity.
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use fnet method. how many obects are being calculated ?
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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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