what is glacioisostacy and what is it doing to earths surface?

Learning Objectives

By the finish of this department, y'all volition exist able to:

• Explain the connection between the constants [latex] K [/latex] and [latex] g [/latex]
• Determine the mass of an astronomical body from free-fall dispatch at its surface
• Depict how the value of [latex] g [/latex] varies due to location and Earth'southward rotation

In this section, we detect how Newton'southward police of gravitation applies at the surface of a planet and how information technology connects with what we learned earlier about free fall. We likewise examine the gravitational furnishings within spherical bodies.

Weight

Call back that the dispatch of a costless-falling object virtually Earth'south surface is approximately [latex] grand=ix.80\,{\text{one thousand/s}}^{2} [/latex]. The force causing this acceleration is chosen the weight of the object, and from Newton's 2nd police, information technology has the value mg. This weight is present regardless of whether the object is in gratuitous fall. Nosotros at present know that this force is the gravitational force betwixt the object and Earth. If we substitute mg for the magnitude of [latex] {\overset{\to }{F}}_{12} [/latex] in Newton's law of universal gravitation, m for [latex] {yard}_{1} [/latex], and [latex] {One thousand}_{\text{Due east}} [/latex] for [latex] {m}_{2} [/latex], we obtain the scalar equation

[latex] mg=M\,\frac{m{Chiliad}_{\text{E}}}{{r}^{2}} [/latex]

where r is the distance betwixt the centers of mass of the object and Globe. The boilerplate radius of Earth is about 6370 km. Hence, for objects within a few kilometers of Earth's surface, nosotros can have [latex] r={R}_{\text{Eastward}} [/latex] ((Figure)). The mass m of the object cancels, leaving

[latex] g=G\frac{{M}_{\text{Due east}}}{{r}^{ii}}. [/latex]

This explains why all masses free fall with the aforementioned acceleration. We accept ignored the fact that Earth besides accelerates toward the falling object, only that is acceptable as long as the mass of Earth is much larger than that of the object.

Figure 13.7 We tin take the distance between the centers of mass of Earth and an object on its surface to exist the radius of Earth, provided that its size is much less than the radius of Earth.

Instance

Masses of Earth and Moon

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

1. Apply the standard values of g, [latex] {R}_{\text{E}} [/latex], and (Effigy) to detect the mass of Earth.
2. Guess the value of yard on the Moon. Utilize the fact that the Moon has a radius of about 1700 km (a value of this accuracy was adamant many centuries agone) and presume information technology has the same boilerplate density equally World, [latex] 5500\,{\text{kg/m}}^{three} [/latex].

Strategy

With the known values of g and [latex] {R}_{\text{E}} [/latex], nosotros can use (Figure) to discover [latex] {Chiliad}_{\text{E}} [/latex]. For the Moon, we utilise the assumption of equal average density to make up one's mind the mass from a ratio of the volumes of Globe and the Moon.

Solution

1. Rearranging (Figure), we take

   [latex] {M}_{\text{E}}=\frac{g{R}_{\text{E}}^{ii}}{G}=\frac{9.80\,{\text{m/south}}^{ii}{(6.37\,×\,{10}^{6}\,\text{m})}^{2}}{6.67\,×\,{10}^{-11}\,\text{N}·{\text{m}}^{2}{\text{/kg}}^{2}}=five.95\,×\,{10}^{24}\,\text{kg.} [/latex]

2. The book of a sphere is proportional to the radius cubed, and so a simple ratio gives us

   [latex] \frac{{G}_{\text{M}}}{{M}_{\text{E}}}=\frac{{R}_{\text{Yard}}^{iii}}{{R}_{\text{E}}^{3}}\to {M}_{\text{1000}}=(\frac{{(one.7\,×\,{10}^{vi}\,\text{k})}^{three}}{{(half-dozen.37\,×\,{10}^{6}\,\text{1000})}^{3}})(v.95\,×\,{10}^{24}\,\text{kg})=1.1\,×\,{10}^{23}\,\text{kg.} [/latex]

   We at present apply (Figure).

   [latex] {g}_{\text{M}}=G\,\frac{{M}_{\text{M}}}{{r}_{\text{1000}}^{2}}=(half dozen.67\,×\,{10}^{-11}\,\text{Due north}·{\text{1000}}^{2}{\text{/kg}}^{2})\frac{(1.1\,×\,{10}^{23}\,\text{kg})}{{(ane.7\,×\,{x}^{6}\,\text{thousand})}^{2}}={ii.5}^{}{\text{grand/s}}^{2} [/latex]

Significance

Every bit soon as Cavendish determined the value of G in 1798, the mass of World could exist calculated. (In fact, that was the ultimate purpose of Cavendish'due south experiment in the commencement place.) The value we calculated for thousand of the Moon is wrong. The average density of the Moon is actually only [latex] 3340\,{\text{kg/1000}}^{3} [/latex] and [latex] chiliad=1.half-dozen\,{\text{m/due south}}^{two} [/latex] at the surface. Newton attempted to mensurate the mass of the Moon by comparing the effect of the Sun on World's ocean tides compared to that of the Moon. His value was a factor of two too pocket-sized. The about 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 exist adamant accurately without going to the Moon. Earth and the Moon orbit about a common center of mass, and careful astronomical measurements tin 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'south center] to [the altitude from the common center of mass to World'south middle].

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

Example

Gravity above Earth's Surface

What is the value of g 400 km higher up World'south surface, where the International Infinite Station is in orbit?

Strategy

Using the value of [latex] {M}_{\text{E}} [/latex] and noting the radius is [latex] r={R}_{\text{East}}+400\,\text{km} [/latex], nosotros employ (Figure) to find g.

From (Figure) we take

[latex] g=G\frac{{M}_{\text{E}}}{{r}^{ii}}=vi.67\,×\,{10}^{-xi}\,\text{North}·{\text{m}}^{2}{\text{/kg}}^{2}\frac{v.96\,×\,{ten}^{24}\,\text{kg}}{{(vi.37\,×\,{10}^{6}+400\,×\,{ten}^{3}\,\text{m})}^{ii}}=8.67\,{\text{m/s}}^{ii}. [/latex]

Significance

Nosotros oftentimes see video of astronauts in infinite stations, evidently weightless. Simply clearly, the force of gravity is acting on them. Comparing the value of chiliad we simply calculated to that on Earth [latex] (ix.eighty\,{\text{m/s}}^{2}) [/latex], nosotros encounter that the astronauts in the International Space Station nevertheless have 88% of their weight. They only announced to exist weightless because they are in free fall. We will come up back to this in Satellite Orbits and Energy.

Check Your Agreement

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

Show Solution

The tallest buildings in the world are all less than ane km. Since k is proportional to the distance squared from World's center, a simple ratio shows that the change in g at ane km higher up World's surface is less than 0.0001%. There would be no demand to consider this in structural pattern.

The Gravitational Field

(Effigy) is a scalar equation, giving the magnitude of the gravitational dispatch equally a function of the altitude from the centre of the mass that causes the acceleration. But we could have retained the vector form for the forcefulness of gravity in (Figure), and written the acceleration in vector form equally

[latex] \overset{\to }{g}=Chiliad\frac{M}{{r}^{2}}\hat{r}. [/latex]

We identify the vector field represented by [latex] \overset{\to }{1000} [/latex] as the gravitational field caused by mass [latex] M [/latex]. We can picture show the field equally shown (Figure). The lines are directed radially inward and are symmetrically distributed about the mass.

Effigy 13.8 A 3-dimensional representation of the gravitational field created by mass [latex] M [/latex]. Annotation that the lines are uniformly distributed in all directions. (The box has been added simply to aid in visualization.)

As is true for any vector field, the direction of [latex] \overset{\to }{k} [/latex] is parallel to the field lines at whatever signal. The strength of [latex] \overset{\to }{g} [/latex] at whatsoever point is inversely proportional to the line spacing. Some other way to state this is that the magnitude of the field in any region is proportional to the number of lines that laissez passer through a unit 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 [latex] {r}^{2} [/latex]. Hence, this picture perfectly represents the changed square law, in addition to indicating the management 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 afterward capacity on electromagnetism.

Apparent Weight: Bookkeeping for Earth'south Rotation

Every bit we saw in Applications of Newton's Laws, objects moving at constant speed in a circle accept a centripetal acceleration directed toward the center of the circle, which means that at that place must be a net force directed toward the center of that circle. Since all objects on the surface of World move through a circumvolve every 24 hours, there must be a net centripetal force on each object directed toward the heart of that circle.

Let's first consider an object of mass yard located at the equator, suspended from a calibration ((Figure)). The scale exerts an upwards strength [latex] {\overset{\to }{F}}_{\text{s}} [/latex] away from Earth'southward heart. This is the reading on the scale, and hence it is the apparent weight of the object. The weight (mg) points toward Earth's center. If Earth were not rotating, the acceleration would be zero and, consequently, the net forcefulness would exist zero, resulting in [latex] {F}_{\text{south}}=mg [/latex]. This would exist the true reading of the weight.

Effigy xiii.nine For a person standing at the equator, the centripetal acceleration [latex] ({a}_{\text{c}}) [/latex] is in the same direction as the force of gravity. At breadth [latex] \lambda [/latex], the angle the betwixt [latex] {a}_{\text{c}} [/latex] and the force of gravity is [latex] \lambda [/latex] and the magnitude of [latex] {a}_{\text{c}} [/latex] decreases with [latex] \text{cos}\lambda [/latex].

With rotation, the sum of these forces must provide the centripetal acceleration, [latex] {a}_{\text{c}} [/latex]. Using Newton's second police, nosotros take

[latex] \sum F={F}_{\text{s}}-mg=m{a}_{\text{c}}\quad \text{where}\quad {a}_{\text{c}}=-\frac{{5}^{2}}{r}. [/latex]

Note that [latex] {a}_{\text{c}} [/latex] points in the same management as the weight; hence, it is negative. The tangential speed 5 is the speed at the equator and r is [latex] {R}_{\text{E}} [/latex]. Nosotros can summate the speed only by noting that objects on the equator travel the circumference of Globe in 24 hours. Instead, allow's use the alternative expression for [latex] {a}_{\text{c}} [/latex] from Motion in Two and Three Dimensions. Recall that the tangential speed is related to the angular speed [latex] (\omega ) [/latex] past [latex] v=r\omega [/latex]. Hence, we accept [latex] {a}_{c}=\text{−}r{\omega }^{two} [/latex]. Past rearranging (Figure) and substituting [latex] r={R}_{\text{E}} [/latex], the apparent weight at the equator is

[latex] {F}_{\text{s}}=m(thou-{R}_{\text{Eastward}}{\omega }^{2}). [/latex]

The angular speed of World everywhere is

[latex] \omega =\frac{ii\pi \,\text{rad}}{24\,\text{hr}\,\,×\,3600\,\text{s/hr}}=7.27\,×\,{x}^{-5}\,\text{rad/s.} [/latex]

Substituting for the values or [latex] {R}_{\text{Eastward}} [/latex] and [latex] \omega [/latex], we have [latex] {R}_{\text{E}}{\omega }^{2}=0.0337\,{\text{k/s}}^{2} [/latex]. This is merely 0.34% of the value of gravity, so it is conspicuously a small correction.

Example

Cypher Apparent Weight

How fast would Earth need to spin for those at the equator to have zero credible weight? How long would the length of the day be?

Strategy

Using (Figure), we tin set the apparent weight ([latex] {F}_{\text{s}} [/latex]) to zero and make up one's mind the centripetal acceleration required. From that, nosotros can find the speed at the equator. The length of day is the time required for 1 complete rotation.

Solution

From (Figure), we take [latex] \sum F={F}_{\text{s}}-mg=m{a}_{\text{c}} [/latex], so setting [latex] {F}_{\text{s}}=0 [/latex], we become [latex] grand={a}_{\text{c}} [/latex]. Using the expression for [latex] {a}_{\text{c}} [/latex], substituting for Earth's radius and the standard value of gravity, we get

[latex] \begin{array}{cc} {a}_{\text{c}}=\frac{{v}^{ii}}{r}=g\hfill \\ v=\sqrt{gr}=\sqrt{(9.80\,{\text{g/southward}}^{2})(6.37\,×\,{10}^{6}\,\text{m})}=7.91\,×\,{10}^{three}\,\text{m/southward}.\hfill \stop{assortment} [/latex]

The period T is the time for one complete rotation. Therefore, the tangential speed is the circumference divided by T, so we take

[latex] \begin{array}{}\\ v=\frac{2\pi r}{T}\hfill \\ T=\frac{ii\pi r}{5}=\frac{2\pi (vi.37\,×\,{10}^{vi}\,\text{chiliad})}{7.91\,×\,{10}^{3}\,\text{grand/south}}=5.06\,×\,{10}^{3}\,\text{s}.\hfill \terminate{array} [/latex]

This is nearly 84 minutes.

Significance

Nosotros will run into later in this chapter that this speed and length of twenty-four hour period would also be the orbital speed and catamenia of a satellite in orbit at Earth'due south surface. While such an orbit would not exist possible almost Earth'south surface due to air resistance, it certainly is possible but a few hundred miles higher up World.

Results Away from the Equator

At the poles, [latex] {a}_{\text{c}}\to 0 [/latex] and [latex] {F}_{\text{s}}=mg [/latex], just as is the case without rotation. At whatsoever other latitude [latex] \lambda [/latex], the situation is more than complicated. The centripetal acceleration is directed toward signal P in the effigy, and the radius becomes [latex] r={R}_{\text{E}}\text{cos}\lambda [/latex]. The vector sum of the weight and [latex] {\overset{\to }{F}}_{\text{s}} [/latex] must betoken toward point P, hence [latex] {\overset{\to }{F}}_{\text{s}} [/latex] no longer points away from the center of Earth. (The difference is small and exaggerated in the figure.) A plumb bob will e'er point along this deviated direction. All buildings are built aligned along this deviated management, not along a radius through the center of Earth. For the tallest buildings, this represents a deviation of a few feet at the peak.

It is besides worth noting that Earth is non a perfect sphere. The interior is partially liquid, and this enhances Earth bulging at the equator due to its rotation. The radius of Earth is almost xxx km greater at the equator compared to the poles. It is left as an practice to compare the strength of gravity at the poles to that at the equator using (Effigy). The difference is comparable to the difference due to rotation and is in the same direction. Plainly, you lot actually tin lose "weight" past moving to the tropics.

Gravity Away from the Surface

Earlier we stated without proof that the law of gravitation applies to spherically symmetrical objects, where the mass of each trunk acts every bit if information technology were at the heart of the trunk. Since (Figure) is derived from (Effigy), it is also valid for symmetrical mass distributions, just both equations are valid only for values of [latex] r\ge {R}_{\text{Due east}} [/latex]. As we saw in (Effigy), at 400 km in a higher place Earth'south surface, where the International Space Station orbits, the value of grand is [latex] 8.67\,{\text{g/southward}}^{2} [/latex]. (Nosotros will run across subsequently that this is also the centripetal dispatch of the ISS.)

For [latex] r<{R}_{\text{Eastward}} [/latex], (Figure) and (Figure) are not valid. Even so, nosotros tin can determine g for these cases using a principle that comes from Gauss'south police force, which is a powerful mathematical tool that we study in more detail afterwards in the course. A consequence of Gauss's constabulary, applied to gravitation, is that only the mass inside r contributes to the gravitational force. Also, that mass, only every bit before, can exist considered to be located at the center. The gravitational effect of the mass outside r has zero net issue.

Two very interesting special cases occur. For a spherical planet with constant density, the mass within r is the density times the book inside r. This mass tin be considered located at the center. Replacing [latex] {Thou}_{\text{E}} [/latex] with simply the mass within r, [latex] Grand=\rho \,×\,(\text{volume of a sphere}) [/latex], and [latex] {R}_{\text{E}} [/latex] with r, (Figure) becomes

[latex] g=Yard\,\frac{{M}_{\text{Due east}}}{{R}_{\text{East}}^{2}}=G\,\frac{\rho (4\text{/}iii\pi {r}^{iii})}{{r}^{two}}=\frac{4}{three}\,M\rho \pi r. [/latex]

The value of g, and hence your weight, decreases linearly equally you lot descend downwardly a hole to the middle of the spherical planet. At the center, you are weightless, as the mass of the planet pulls equally in all directions. Actually, Earth's density is not constant, nor is Globe solid throughout. (Effigy) shows the profile of g if Earth had constant density and the more probable profile based upon estimates of density derived from seismic data.

Figure 13.10 For [latex] r<{R}_{\text{Due east}} [/latex], the value of k for the instance of constant density is the straight green line. The bluish line from the PREM (Preliminary Reference Earth Model) is probably closer to the actual profile for one thousand.

The second interesting case concerns living on a spherical shell planet. This scenario has been proposed in many scientific discipline fiction stories. Ignoring pregnant engineering issues, the shell could be constructed with a desired radius and total mass, such that chiliad at the surface is the aforementioned as Earth's. Tin you lot guess what happens once yous descend in an elevator to the inside of the shell, where at that place is no mass betwixt you and the heart? What benefits would this provide for traveling great distances from one point on the sphere to some other? And finally, what effect would at that place be if the planet was spinning?

Summary

• The weight of an object is the gravitational allure betwixt Globe and the object.
• The gravitational field is represented as lines that indicate the direction of the gravitational force; the line spacing indicates the strength of the field.
• Credible weight differs from actual weight due to the dispatch of the object.

Conceptual Questions

Must engineers take Earth's rotation into account when constructing very alpine buildings at any location other than the equator or very virtually the poles?

Show Solution

The centripetal dispatch is not directed along the gravitational strength and therefore the right line of the building (i.due east., the plumb bob line) is not directed towards the middle of Earth. But engineers utilize either a plumb bob or a transit, both of which respond to both the management of gravity and acceleration. No special consideration for their location on World need be made.

Problems

(a) Calculate Earth's mass given the acceleration due to gravity at the North Pole is measured to be [latex] 9.832\,{\text{m/s}}^{2} [/latex] and the radius of the Earth at the pole is 6356 km. (b) Compare this with the NASA'south Globe Fact Sheet value of [latex] 5.9726\,×\,{10}^{24}\,\text{kg} [/latex].

(a) What is the acceleration due to gravity on the surface of the Moon? (b) On the surface of Mars? The mass of Mars is [latex] vi.418\,×\,{10}^{23}\,\text{kg} [/latex] and its radius is [latex] 3.38\,×\,{x}^{6}\,\text{m} [/latex].

Bear witness Solution

a. [latex] one.62\,{\text{m/s}}^{2} [/latex]; b. [latex] 3.75\,{\text{yard/s}}^{2} [/latex]

(a) Calculate the acceleration due to gravity on the surface of the Sunday. (b) By what gene would your weight increase if you lot could stand up on the Sun? (Never listen that you cannot.)

The mass of a particle is 15 kg. (a) What is its weight on Earth? (b) What is its weight on the Moon? (c) What is its mass on the Moon? (d) What is its weight in outer space far from whatsoever celestial trunk? (e) What is its mass at this bespeak?

Show Solution

a. 147 North; b. 25.v N; c. 15 kg; d. 0; e. 15 kg

On a planet whose radius is [latex] 1.2\,×\,{x}^{vii}\,\text{thou} [/latex], the acceleration due to gravity is [latex] 18\,{\text{1000/s}}^{2} [/latex]. What is the mass of the planet?

The mean bore of the planet Saturn is [latex] i.two\,×\,{ten}^{eight}\,\text{k} [/latex], and its mean mass density is [latex] 0.69\,{\text{g/cm}}^{3} [/latex]. Find the dispatch due to gravity at Saturn's surface.

Evidence Solution

[latex] 12\,{\text{m/southward}}^{2} [/latex]

The mean bore of the planet Mercury is [latex] 4.88\,×\,{10}^{half dozen}\,\text{chiliad} [/latex], and the acceleration due to gravity at its surface is [latex] 3.78\,{\text{m/due south}}^{ii} [/latex]. Approximate the mass of this planet.

The acceleration due to gravity on the surface of a planet is iii times as large as it is on the surface of Earth. The mass density of the planet is known to exist twice that of Earth. What is the radius of this planet in terms of Globe'southward radius?

Testify Solution

[latex] (3\text{/}2){R}_{\text{East}} [/latex]

A body on the surface of a planet with the same radius as Earth's weighs 10 times more than than it does on Earth. What is the mass of this planet in terms of Earth'south mass?

Glossary

apparent weight

reading of the weight of an object on a calibration that does not business relationship for acceleration

gravitational field

vector field that surrounds the mass creating the field; the field is represented by field lines, in which the management of the field is tangent to the lines, and the magnitude (or field forcefulness) is inversely proportional to the spacing of the lines; other masses respond to this field

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