Summary of Equations
Two points in space, A and B.
Distances from the center of the source of gravitation are RA and RB.
a = acceleration, W = weight.
aB/aA = (RA/RB)2
WB/WA = (RA/RB)2
Problem 1
Take point A as the surface of the earth. Distance of object A from the center of the earth equals the radius of the earth, RA = 6400 km.
Weight at earth's surface is WA = 800 N.
(a) Point B is 6400 km above the surface, RB = 6400 + 6400 = 12,800 km.
WB/WA = (RA/RB)2
WB/800 = (RA/RB)2 = (6400/12800)2 = (1/2)2 = 1/4
WB = (800)(1/4) = 200 N.
(b) RA = 6400 km, WA = 800 N, as in part (a).
RB = 6400 + 12,800 = 19,200
WB/800 = (RA/RB)2 = (6400/19200)2 = (1/3)2 = 1/9
WB = (800)(1/9) = 88.9 N
(c) RA = 6400 km, WA = 800 N, as in part (a).
RB = 6400 + 19,200 = 19,200 = 25,600 km
WB/800 = (RA/RB)2 = (6400/25600)2 = (1/4)2 = 1/16
WB = (800)(1/16) = 50 N
Problem 2
Distances are the same as in problem 1.
aA = acceleration at the earth's surface, aA = g = 9.8 m/s2.
(a) aB/aA = (RA/RB)2
aB/9.8 = 1/4
aB = (9.8)(1/4) = 2.45 m/s2.
(b) aB = (9.8)(1/9) = 1.09 m/s2.
(c) aB = (9.8)(1/16) = 0.613 m/s2.
Problem 3
Here point A is the surface of the moon, RA is the distance from the center of the moon to the surface, i.e. 1600 km, WA = 100 N
(a) RB = 1600 + 3200 = 4800 km
WB/WA = (RA/RB)2
WB/100 = (1600/4800)2 = (1/3)2 = 1/9
WB = (100)(1/9) = 11.1 N
(b) RB = 1600 + 5000 = 6600 km
WB/100 = (1600/6600)2 = (0.242)2 = 0.0586
(Note: If you don't round off the 0.242 before you square it, you'll get 0.0588 here instead of 0.0586. We won't be concerned with small differences like this.)
WB = (100)(0.0586) = 5.86 N
Problem 4
Distances are the same as in problem 3.
aA = 1.7 m/s2
(a) aB/aA = (RA/RB)2
aB/1.7 = 1/9
aB = (1.7)(1/9) = 0.189 m/s2
(b) aB/1.7 = (1600/6600)2 = (0.242)2 = 0.0586
aB = (1.7)(0.0586) = 0.0996 m/s2
Problem 5
For surface of earth, RA = 6400 km.
WA = 618 N
In the satellite (point B), distance from center of earth is 6400 + 200 = 6600 km.
WB/WA = (RA/RB)2
WB/618 = (6400/6600)2 = (0.970)2 = 0.941
WB = (618)(0.941) = 582 N
Problem 6 and 7
Solutions are given in the link to the
answers.
Problem 8
You can write a proportionality, comparing the new satellite to the known satellite, Io (or any other satellite with known properties). For Io we have distance = 422,000 meters, acceleration = 0.7124 m/s2. Therefore,
asatellite/0.7124 = (422,000/1,120,000)2 = (0.377)2 = 0.142
asatellite = 0.7124 X 0.142 = 0.101 m/s2
Problem 9
If the asteroid's distance from the sun is twice the distance of Mars, the asteroid's acceleration must be one-fourth the acceleration of Mars. The one-fourth comes from the square of one-half. The acceleration of Mars is 0.00256 m/s2.
Acceleration of asteroid = (1/4)X(0.00256) = 0.00064 m/s2.
Problem 10
We compare the asteroid's acceleration to the acceleration of Mars:
aast/aMars = (RMars/Rast)2
aMars = 0.00256 m/s2
RMars = 2.28 X 1011 meters.
aast/0.00256 = (2.28 X 1011/5.0 X 1011)2 = (0.456)2 = 0.208
aast = 0.00256 X 0.208 = 0.00053 m/s2