http://www.ck12.org Chapter 9. Newton’s Universal Law of Gravity
Gmmme
R^2
=mm
4 π^2 R
T^2
and solve forT^2 →T^2 =
4 π^2 R^3
Gme.
The derived expression, not coincidentally, looks a lot like Kepler’s Third Law (applied to circular orbits), where the
constantkmentioned above is equal isk=^4 π
2
Gme. The fact that the Universal Law of Gravity yielded Kepler’s Third
Law gave a good deal of support to its validity. It also explained that Kepler’s Third Law had the form it did as a
result of an inverse square force law.
It is important to note that the mass in the denominator ofkis the mass of the central body. In the work above, it is
the mass of the Earth. The mass of the orbiting body does not enter the equation. It is beyond the scope of this book,
but a more general statement of Kepler’s Third Law does use both masses.
http://m.wolframalpha.com/input/?i=Kepler%27s+third+law%2C+4+solar+masses%2C+5+Earth+masses%2C+2.5+A
U
Illustrative Example 1
1a. Use the Universal Law of Gravity to find the period of the Moon about the Earth.
Note,G= 6. 67 × 10 −^11 N∗m
2
kg^2 , the distance between the centers of the Earth and Moon isR=^3.^84 ×^10
(^5) kmand the
mass of Earth isme= 5. 97 × 1024 kg.
Answer:
From the above we haveT^2 =^4 π
2
GmeR
(^3). We must remember that the SI system of units uses meters and seconds, and
therefore we must convertR= 3. 84 × 105 km→R= 3. 84 × 108 m.
Substituting into the above equation we have:
T^2 =
4 π^2
GmeR^3 →
4 π^2
( 6. 67 × 10 −^11 )( 5. 97 × 1024 )( 3. 84 × 108 )^3 = 5. 614 × 1012
→T^2 = 5. 614 × 1012 →T= 2. 369 × 106 sLet us go one step further and represent this result in days instead of seconds. One day has 86,400 seconds, therefore
2. 369 × 106 s
86 , (^400) ds =^27.^42 days. The actual period of the Moon is closer to 27.3 days. Still, a good enough result to assure
us that the Universal Law of Gravity yields correct results!
1b. Show that the gravitational accelerationgvaries as the inverse square with the distance from the center of the
Earth.
Answer:
For an object of massmat rest upon the surface of the Earth, the object experiences a forceFdue to the Earth, of
F=mg. But forceFmust be the same as that given by the Universal Law of Gravity, namelyF=GmR 2 em, whereRis
the radius of the Earth. Thus,mg=GmR 2 em→g=GmR 2 e. This is why in all of our work so far we have always made the
statement thatg= 9. (^81) sm 2 near the Earth’s surface. Try substituting the values forG,me,andRin the above equation
forgto show thatg= 9. 81 ms 2.
Check Your Understanding
What gravitational acceleration does an object experience at the instant it is released at one Earth radius above the
surface of the Earth? SeeFigure9.8.