9.2. Newton’s Universal Law of Gravity http://www.ck12.org
FIGURE 9.8
Answer:
Sinceg=Gmr 2 e, we substituter= 2 R
g=
Gme
r^2=
Gme
( 2 R)^2=
Gme
4 R^2=
1
4
Gme
R^2=
1
4
g=1
4
( 9. 81
m
s^2) = 2. 45
m
s^2Remember that the distanceris measured from the center of the Earth!
Illustrative Example 2
Another support for the Universal Law of Gravity comes from computing the acceleration of the Moon toward the
Earth. The distance from the center of the Moon to center of the Earth is 60.27 Earth radii(Re), so the acceleration of
the Moon toward the Earth should be( 60.^127 ) 2 ge, wherege= 9. 81 ms 2 the acceleration of gravity at the Earth’s surface.
a. Find the acceleration of the Moon(gat moon)relative to the Earth using the equationa=v
2
r.Answer: We have already shown thata=^4 π
(^2) R
T^2 , that the period of the Moon about the Earth in seconds isT=
2. 369 × 106 s, and the distance between their centers in meters isR= 3. 84 × 108 m. Substituting these values into
a=^4 π
(^2) R
T^2 , we havea=
4 π^2 ( 3. 84 × 108 m)
( 2. 369 × 106 s)^2 =^0.^00270
m
s^2
b. Find the acceleration of the Moon relative to the Earth if the distance between their centers is 60. 27 Reusing
Newton’s Universal Law of Gravity. As discussed above, the Universal Law of Gravity gives the acceleration of the
Moon at a distance of 60. 27 Reas( 601. 27 ) 2 g.
Answer:
1
( 60. 27 )^2 g=
9. 81
( 60. 27 )^2 =^0.^00270
m
s^2
A more complete solution is: