Procedimientos

One very common (and practical) type of gravity problem in astronomy is to determine the force of gravity between a celestial body (such as a moon, planet, or star) and an object on the surface³of that body. According to Newton’s Law of Gravity, that force depends on the mass of the celestial body, the mass of the object on the surface, and the distance from the center of the body to the center of the object.

To understand the relevant distance, consider the case of a person standing on the surface of a planet, as shown in Figure 2.3. As suggested by this figure, the distance from the center of the planet to the center of the person is very well approximated by the radius of the planet (since the planet’s radius is typically thousands of kilometers and the person’s height is 2 meters or less). So in this case the “R” term in the denominator of Newton’s Law of Gravity turns out to be the radius of the body to a very good approximation.

This means that if you (with mass myou) are standing on the surface of a planet (with mass mplanet), the force of gravity (Fg) between you and the planet is

Fg= Gmyoumplanet

R²_planet .

Distance from center of planet to center of person

Radius of planet

Person

Center of planet

Figure 2.3 Radius of planet and center-to-center distance.

3 Some objects, such as stars and gas-giant planets, lack a well-defined surface. For stars, the

“surface” is often defined as the photosphere of the star (that’s the layer from which the star’s light radiates), although no solid surface exists at that location. Likewise, for gas-giant planets, the “surface” level is sometimes associated with a layer in the planet’s atmosphere.

2.1 Newton’s Law of Gravity 47

Here’s an example:

Example: Find the force of gravity of the Earth on a person with a mass of 100 kg standing on the Earth’s surface.

In this case, the only number you’re given is the mass of the person (100 kg), which you can call m1. But you’re also told that the person is standing on the surface of the Earth, which means that you can look up the mass of the second object (the Earth), which you can call m2, and the distance between the person and the center of the Earth (R), which is essentially the radius of the Earth. In any comprehensive astronomy text, you should be able to find⁴the mass of the Earth as about 6× 10²⁴kg and the radius of the Earth as approximately 6,378 km. So, you have the values of two masses (the person and the Earth) as well as the distance between their centers.

Clearly, the relationship between two masses, the distance between their centers, and the force of gravity between them is provided by Newton’s Law of Gravity. So you can solve this problem by using the gravity equation, but first you’re going to have to make sure that the variables have the units required by that equation. You have the mass of the person and the mass of the Earth in kilograms, as required, but the distance between their cen-ters is given in kilomecen-ters rather than mecen-ters. That’s an easy conversion (see Section 1.1 if you need help with unit conversions); multiplying 6,378 km by the conversion factor of 1,000 meters per kilometer gives you the value of R:

6,378,000 m, or 6.378×10⁶m (see Section 1.4 if you need help with scientific notation).

Inserting these values along with the constant G into the gravity equation gives

which is the gravitational force between the Earth and a 100-kg person standing on the Earth’s surface. To put this into more-familiar terms, you can convert the units of this result from newtons to pounds by using the conversion factor

4 If you don’t know where to find it, use the index.

48 Gravity

of 1 lb↔ 4.45 N. This means that the force of Earth’s gravity in pounds on a 100-kg person is

Fg= (983.8 N) 1 lb

4.45 N= 221 lb.

As always, you should ask yourself whether this answer makes sense. And it does! Since 1 kg “equals” 2.2 lb, it makes sense that 100 kg “equals” about 220 lb. But can 1 kg (which is an amount of mass) really “equal” 2.2 lb (which is a force)? Clearly not, which is why we put quotation marks around the word

“equals” in the previous sentences. The reason it’s commonly said that 1 kg equals 2.2 lb is that on the surface of the Earth the force of Earth’s gravity on a mass of 1 kg is about 2.2 lb. But take that 1-kg mass just about anywhere else in the Universe, and the force of gravity on it will not be 2.2 lb. To verify that, you can use the gravity equation to find the force of gravity produced by Earth’s Moon on a mass on the surface of the Moon.

Example: Find the force of the Moon’s gravity on a 1-kg object on the surface of the Moon.

The Moon’s mass is about 7.35 × 10²²kg, and its radius is about 1,737 km, so the gravity equation gives

Fg= Gm1m2

R² (2.1)

Fg=



6.67 × 10⁻¹¹N m² kg²

(1 kg)(7.35 × 10²²kg) (1.737 × 10⁶m)²

= 1.62 N,

which is about 0.37 lb. This result means that 1 kg does not “equal” 2.2 lb on the surface of the Moon. Instead, on the surface of the Moon, a mass of 1 kg weighs only about¹₃ lb.

Since 0.37 is very close to one-sixth of 2.2, it’s frequently said that you would weigh only about one-sixth as much on the Moon as on Earth, or “the Moon’s gravity is one-sixth of Earth’s gravity”.

It is precisely this kind of comparison that is easily done using the ratio method described in Chapter 1. Whenever you’re asked to compare quantities (often through questions such as “How many times bigger. . .?” or “How much stronger. . .?”), you should consider using the ratio method. But even when you’re trying to find an absolute quantity such as the force of gravity on a certain planet, as long as you have a reference value (such as the force of gravity on Earth), you may still be able to save a lot of time and effort – and minimize what you have to plug into your calculator, thus reducing chances for errors – by using ratios where possible.

2.1 Newton’s Law of Gravity 49

You can find a detailed description of and motivation for the ratio method in Section 1.2, but here’s a quick summary of how to apply it to an example gravity problem.

Example: Compare the force of Earth’s gravity on an object on Earth’s sur-face with the force of the Moon’s gravity on the same object on the sursur-face of the Moon.

If you use the ratio method, it’s not necessary to go through the process of finding the value of Fgon the Earth, then finding the value of Fgon the Moon, and then dividing one by the other (if you’re wondering why it’s often better to compare quantities in astronomy by dividing rather than by subtracting, that’s also discussed in Section 1.2).

You can certainly get the correct answer that way, but you’re taking a lot of steps that can be avoided by realizing that quantities of the mass of the object (call it m1) and G will be exactly the same in your calculations for both locations (Earth and Moon).

By not plugging in values until the very end, you’ll see that m1and G will cancel out.

This may be a little more clear if you consider what’s going to happen when you take that crucial “compare by dividing” step. If you call the force of gravity on Earth Fg,Earthand the force of gravity on the Moon Fg,Moon, dividing will look like this:

Now plugging in values for the masses of Earth and Moon, and the distances between their centers and surfaces (which are equal to their radii), gives

Fg,Moon

This is the same result you obtained earlier using the absolute method to cal-culate the force of the Moon’s gravity on a 1-kg mass, convert that force from newtons to pounds and finally to compare it to the weight of a 1-kg mass on Earth (2.2 lb). But in the ratio approach you didn’t have to enter a value for m1

because it cancelled. In other words, the result is independent of mass m1– the

50 Gravity

Moon’s gravity is one-sixth of Earth’s gravity for an object of any mass. Fur-thermore, notice that this time you didn’t have to do a unit conversion to get distance into units of meters. Since you had both distances in kilometers, when you divided them the units cancelled. So with the ratio method the quantities can be in any units, as long as the units are the same in the two scenarios you are comparing.

The simplicity of this result dramatically illustrates the power of the ratio method. To compare the Moon’s gravity to the Earth’s, just look at Eq. 2.2.

This equation says that to compare Fg,Moonto Fg,Earth, you can simply divide the mass of the Moon by the mass of the Earth and multiply that result by the square of the ratio of the Earth’s radius to the Moon’s radius. In terms of calculator operations, the four number entries, two divides, one squaring, and one multiply (8 total entries) of the ratio method replace the ten number entries, five divides, two squarings, and four multiplies (21 total entries) of the absolute method, and that’s not counting unit conversions.

The power of the ratio method is even more compelling in cases in which you’re already given values in ratio form, as you can see in the following example.

Example: Compare Earth’s surface gravity to that of Jupiter, whose radius is 11.2 times the radius of the Earth and whose mass is 318 times the mass of Earth.

In such a problem, there’s no need to calculate the force of gravity on a certain mass (m1) on Jupiter and on Earth and then divide one result by the other. Just as in the previous example, m1and G will cancel so you can simply start with the Jupiter/Earth version of Eq, 2.2:

Fg,Jupiter or Fg,Jupiter = 2.5 Fg,Earth. Translating this mathematical result into a sen-tence, this means that the force of gravity on the “surface” of Jupiter is 2.5 times stronger than the force of gravity on the surface of Earth (for help interpreting ratio answers, see Section 1.2.4).

If it seems strange to you that the force of gravity on the surface of a planet that’s over 300 times more massive than Earth is only 2.5 times greater than the force of gravity on the surface of the Earth, remember the ¹

R²-term in the denominator of Newton’s Law of Gravity. Because Jupiter is much larger than Earth, the distance from the “surface” to the center of Jupiter is much greater than the distance from the surface to the center of Earth, and this greater distance partially compensates for the greater mass of Jupiter.

2.2 Newton’s Laws of Motion 51

You can get some practice doing gravity problems by working through the following exercise, and you’ll find several gravity problems in the problem set at the end of this chapter.

Exercise 2.2. Calculate the force of gravity (that is, the weight) in newtons