PhysicsGravitation: Force
Science and Mathematics Education Research Group
Supported by UBC Teaching and Learning Enhancement Fund 2012-2013
F ACULTY OF EDUCATION F ACULTY OF EDUCATION
Department of Curriculum and Pedagogy
FACULTY OF EDUCATION
Question TitleQuestion TitlePlanetoids
A
C
B
D
Question Title
A. The force on the smaller mass is greater
B. The force on the larger mass is greater
C. They are equal
D. No idea
Two planetoids of unequal mass are separated by an unknown distance. What can be said of the gravitational force on each mass?
Question TitlePlanetoids I
Comments
Answer: C
Justification:
1. The equation for the gravitational force one planet exerts on another is given below. The equation does not change when considering the force 1 exerts on 2, or the force 2 exerts on 1. Therefore, the two forces are equal.
2. Another way of thinking about this problem: By Newton’s third law the force the smaller mass exerts on the larger mass is equal to the force the larger mass exerts on the smaller mass.
CommentsSolution
1 2 2 11 2 2 12 2
and on on
Gmm Gm mF F
r r
Question Title
If mC = mB, and the distances are as shown (with each of the lines equal to r and perpendicular), what is the magnitude of the force A exerts on C as compared to the force A exerts on B?
Question TitlePlanetoids II
. 2
.
. 2
. 2
. None of the above
AonB
AonB
AonB
AonB
A F
B F
FC
FD
E
A C
B
r r
Comments
Answer: D
Justification: The gravitational force is proportional to the inverse of the distance squared ( ). The distances between A and B, and B and C are both r. By the Pythagorean theorem, the distance between A and C is . Therefore:
Thus, the force between A and C is 2 times weaker than the force between A and B.
CommentsSolution
2
1
rF
2r
2 2 2 (Remember )
2 2 22
A C A C AonBA BAonC B C
m m m m Fm mF G G G m m
r rr
Question Title
If mC is equal to mB, and is twice mA, and the distances are as shown (with each of the lines equal to r and perpendicular), what is the magnitude of the net force on B, in terms of the force A exerts on B?
Question TitlePlanetoids III
A C
B
r r
. 2
.
. 3
. 2
. 5
AonB
AonB
AonB
AonB
AonB
A F
B F
C F
FD
E F
Comments
Answer: E
Justification: Since the gravitational force scales linearly with mass, the force C exerts on B is twice as large as the force A exerts on B:
CommentsSolution
FAonB
FConB = 2FAonB
FnetB
The two forces FAonB and FConB are perpendicular to each other, and can be added using Pythagorean Theorem:
2 2 2
22 2 (Remember 2 )C B A B A B
ConB AonB C A
m m m m m mF G G G F m m
r r r
2 2 2 2
2
2
5 = 5
net AonB ConB AonB AonB
net AonB AonB
F F F F F
F F F
Question Title
If mC is equal to mB and mD, and is twice mA, and the distances are as shown (with each of the lines equal to r and perpendicular), what is the magnitude of the net force on A?
Question TitlePlanetoids IV
A
C
B
D
above theof None .
21 .
2
11 .
2
5 .
2
11 .
E
FD
FC
FB
FA
AonB
AonB
AonB
AonB
r
r
Comments
Answer: A
Justification: In question 2 we found that .
Because D has the same mass as C and is located the same distance from A as C, the magnitudes of the forces exerted by each are the same, . The forces are also perpendicular to each other.
Thus the net force exerted on A by C and D can be calculated using Pythagorean theorem.
CommentsSolution
FConA FDonA
FBonA
2AonB
ConA
FF
DonAConA FF
2 2 2
22 2 4 2BonA BonA BonA BonAF F F F
Adding FBonA
1
2
1
2BonABonA
BonA FFF
Note: According to Newton’s third law: FBonA = FAonB (equal magnitudes)
Question Title
If mC is equal to mB and mD, and is twice mA, and the distances are as shown (with each of the lines equal to r and perpendicular), what is the magnitude of the net force on B?
Question TitlePlanetoids V
A
C
B
D
AonB
AonB
AonB
AonB
AonB
FE
FD
FC
FB
FA
21 .
.5
10 .
2
5 .
2
11 .
r
r
Comments
Answer: D
Justification: Because mC= mD, the forces exerted by C and D on B are equal in magnitude and opposite in direction. Therefore, they cancel each other out.
We are then left with the force that A exerts on B, FAonB.
CommentsSolution
