In the previous section, we saw the magnitude of direction of the gravitational force of attraction . We also saw a solved example. In this section we will see a few more solved examples
Solved example 8.8
Find the unit of the universal gravitational constant G
Solution:
• We have: |→FG|=Gm1m2|→r|2
• So we get: G=|→r|2|→FG|m1m2
• Thus the unit is: Nm2kg−2
Solved example 8.9
Two heavy bodies of masses 40 kg and 60 kg attract each other with a force of 4 × 10-5 N. If G is 6 × 10-11 N m2 kg-2, find the distance between them
Solution:
• We have: |→FG|=Gm1m2|→r|2
• Substituting the given values, we get: 4×10−5(N)=6×10−11(Nm2kg−2)(40(kg)×60(kg)|→r|2)
⇒4×10−5=(1.44×10−7(m2)|→r|2)
⇒|→r|2=(1.44×10−7(m2)4×10−5)=0.0036(m2)
⇒|→r|=0.06m
Solved example 8.10
Two bodies A and B of masses M and 4M respectively, are placed at a distance of 2.73 m apart. Another body C of mass m is to be placed in between A and B in such a way that, the net force on C is zero. What is the distance of C from A
Solution:
1. Fig.8.14 (a) below shows the 3 bodies in a line
Let C be at a distance of x m from A
2. Fig.b shows the free body diagram of C
(i) We have: →FG(A,C)=−[GM×m|→rA,C|3]→rA,C
⇒→FG(A,C)=−[GM×mx3](x)(−ˆi)
⇒→FG(A,C)=[GMmx2]ˆi
(ii) We have: →FG(B,C)=−[Gm×4M|→rB,C|3]→rB,C
⇒→FG(A,C)=−[Gm×4M(2.73−x)3](2.73−x)ˆi
⇒→FG(A,C)=−[4GmM(2.73−x)2]ˆi
3. Net force = →FG(A,C)+→FG(B,C)
⇒ Net force = [GMmx2]ˆi+(−[4GmM(2.73−x)2]ˆi)
4. But net force must be zero. So we get:
[GMmx2]ˆi=[4GmM(2.73−x)2]ˆi
• We have an equality of two vectors. So their magnitudes must be the same. We can write:
[GMmx2]=[4GmM(2.73−x)2]
⇒[1x2]=[4(2.73−x)2]
⇒(2.73−x)2=4x2
⇒ (2.73-x) = 2x (Taking roots on both sides)
⇒ 2.73 = 3x
⇒ x = 0.91 m
Solved example 8.11
Two bodies A and B have equal masses. The gravitational force of attraction between them is →FG. 25% of the mass of A is transferred to B. What is the new force of attraction
Solution:
1. Let the initial masses of the two bodies be m each
• Then we can write: →FG=−[Gm2|→r|3]→r
• Where →r is the distance vector between A and B
2. Final mass of A = (m-0.25m) = 0.75m
• Final mass of B = (m+0.25m) = 1.25m
3. New force →FG(new)=−[G(0.75m)(1.25m)|→r|3]→r=−[0.9375Gm2|→r|3]→r
4. Dividing (3) by (1), we get:
→FG(new)→FG=(−[0.9375Gm2|→r|3]→r)÷(−[Gm2|→r|3]→r)=0.9375
⇒→FG(new)=0.9375→FG
Solved example 8.12
Two bodies A and B have masses m and M respectively. The gravitational force of attraction between them is →F. A third body C is placed in contact with A. The body C has a mass of 2m. What is the force on B due to A? What is the total force on B ?
Solution:
Part (a):
• Initially, there are only two masses A and B
• At that time, the force is given to be →F
• We can write:
→F=→FG(A,B)=−[GmM|→r|3]→r
• Where →r is the distance vector between A and B
Part (b):
• When C is placed near A, the force on B due to C is given by:
→FG(C,B)=−[G×2m×M|→r|3]→r
• Taking ratios, we get:
→FG(C,B)→FG(A,B)=→FG(C,B)→F=(−[G×2m×M|→r|3]→r)÷(−[G×m×M|→r|3]→r)
⇒→FG(C,B)→F=2
⇒→FG(C,B)=2→F
• Thus we can write:
Total force on B = →FG(A,B)+→FG(C,B)=→F+2→F=3→F
Solved example 8.13
In fig.8.15(a) below, three equal masses of m kg each are fixed at three corners A, B and D of the square of side a
What is the force acting on a mass 1 kg placed at the fourth corner C ?
Solution:
1. Any square will lie on a plane. So this is a 2D problem. We will need the x and y axes only. There is no need for the z-axis
• Assume that, the origin of the coordinate axes is at A. This is shown in fig.b
2. For this square:
(i) C will be equidistant from B and D
C will be at a distance of (2)a from A
3. Now we can begin the calculations:
• The distance vectors are shown in fig.8.16(a) below:
• The free body diagram of the 1 kg mass at C is shown in fig.16(b)
4. Force due to mass at B
(i) →FG(B,C)=−[Gm×1|→rB,C|3]→rB,C
⇒→FG(B,C)=−[Gma3]→rB,C
(ii) So our next task is to write →rB,C in component form
• To reach C from B:
♦ We have no horizontal distance to travel
♦ We travel only a distance BC vertically up
• So the components of →rB,C are (0)ˆi and aˆj
• We can write: →rB,C=aˆj
(iii) So the result in (i) becomes: →FG(B,C)=−[Gma3]aˆj
5. Force due to mass at D
(i) →FG(D,C)=−[Gm×1|→rD,C|3]→rD,C
⇒→FG(D,C)=−[Gma3]→rD,C
(ii) So our next task is to write →rD,C in component form
• To reach C from D:
♦ We have no vertical distance to travel
♦ We travel only a distance DC horizontally to the right
• So the components of →rD,C are (a)ˆi and 0ˆj
• We can write: →rD,C=aˆi
(iii) So the result in (i) becomes: →FG(D,C)=−[Gma3]aˆi
6. Force due to mass at A:
(i) →FG(A,C)=−[Gm×1|→rA,C|3]→rA,C
⇒→FG(A,C)=−[Gm(√2a)3]→rA,C
(ii) So our next task is to write →rA,C in component form
• To reach C from A:
♦ First we travel a distance AB horizontally to the right
♦ Then we travel a distance BC vertically up
• So the components of →rA,C are aˆi and aˆj
• We can write: →rA,C=aˆi+aˆj
(iii) So the result in (i) becomes:
→FG(A,C)=−[Gm(√2a)3](aˆi+aˆj)
⇒→FG(A,C)=(−[Gm(√2a)3]aˆi)+(−[Gm(√2a)3]aˆj)
7. When we add the results in (4), (5) and (6), we get the net force experienced by the mass at C
So we can write:
The net force experienced by the mass at C
= →FG(B,C)+→FG(D,C)+→FG(A,C)
= (−[Gma3]aˆj)+(−[Gma3]aˆi)+(−[Gm(√2a)3]aˆi)+(−[Gm(√2a)3]aˆj)
(i) First we will add the horizontal terms:
(−[Gma3]aˆi)+(−[Gm(√2a)3]aˆi)
= −[Gma3](aˆi+[12√2]aˆi)
= −[Gma2](1+12√2)ˆi
(ii) Next we add the vertical terms:
(−[Gma3]aˆj)+(−[Gm(√2a)3]aˆj)
= −[Gma3](aˆj+[12√2]aˆj)
= −[Gma2](1+12√2)ˆj
8. We can write:
■ The resultant force on C has two components:
• The horizontal component is: −[Gma2](1+12√2)ˆi
• The vertical component is: −[Gma2](1+12√2)ˆj
9. Note that, the magnitudes of both components are the same
So we can write:
■ The resultant force on C has two components:
• The horizontal component is: −pˆi
• The vertical component is: −pˆj
Where p=[Gma2](1+12√2)
This is shown in fig.8.17(a) below:
10. We have two vectors
♦ The magnitudes of the two vectors are equal
♦ The two vectors are perpendicular to each other
• So the magnitude of the resultant vector will be given by:
√p2+p2=√2p2=√2p (Details here)
= √2[Gma2](1+12√2)
= [Gma2](√2+12)
• Also, the resultant will be making an angle of 45o with the horizontal. This is shown in fig.8.17(b)
• In (6) we wrote: →rA,C=aˆi+aˆj
• We can use trigonometry and write:
→rA,C=(acos45)ˆi+(asin45)ˆj
• The reader is advised to write all steps using this alternate method and prove that, the resultant force on the mass at C is the same as that obtained above
Solved example 8.8
Find the unit of the universal gravitational constant G
Solution:
• We have: |→FG|=Gm1m2|→r|2
• So we get: G=|→r|2|→FG|m1m2
• Thus the unit is: Nm2kg−2
Solved example 8.9
Two heavy bodies of masses 40 kg and 60 kg attract each other with a force of 4 × 10-5 N. If G is 6 × 10-11 N m2 kg-2, find the distance between them
Solution:
• We have: |→FG|=Gm1m2|→r|2
• Substituting the given values, we get: 4×10−5(N)=6×10−11(Nm2kg−2)(40(kg)×60(kg)|→r|2)
⇒4×10−5=(1.44×10−7(m2)|→r|2)
⇒|→r|2=(1.44×10−7(m2)4×10−5)=0.0036(m2)
⇒|→r|=0.06m
Solved example 8.10
Two bodies A and B of masses M and 4M respectively, are placed at a distance of 2.73 m apart. Another body C of mass m is to be placed in between A and B in such a way that, the net force on C is zero. What is the distance of C from A
Solution:
1. Fig.8.14 (a) below shows the 3 bodies in a line
Let C be at a distance of x m from A
![]() |
Fig.8.14 |
(i) We have: →FG(A,C)=−[GM×m|→rA,C|3]→rA,C
⇒→FG(A,C)=−[GM×mx3](x)(−ˆi)
⇒→FG(A,C)=[GMmx2]ˆi
(ii) We have: →FG(B,C)=−[Gm×4M|→rB,C|3]→rB,C
⇒→FG(A,C)=−[Gm×4M(2.73−x)3](2.73−x)ˆi
⇒→FG(A,C)=−[4GmM(2.73−x)2]ˆi
3. Net force = →FG(A,C)+→FG(B,C)
⇒ Net force = [GMmx2]ˆi+(−[4GmM(2.73−x)2]ˆi)
4. But net force must be zero. So we get:
[GMmx2]ˆi=[4GmM(2.73−x)2]ˆi
• We have an equality of two vectors. So their magnitudes must be the same. We can write:
[GMmx2]=[4GmM(2.73−x)2]
⇒[1x2]=[4(2.73−x)2]
⇒(2.73−x)2=4x2
⇒ (2.73-x) = 2x (Taking roots on both sides)
⇒ 2.73 = 3x
⇒ x = 0.91 m
Solved example 8.11
Two bodies A and B have equal masses. The gravitational force of attraction between them is →FG. 25% of the mass of A is transferred to B. What is the new force of attraction
Solution:
1. Let the initial masses of the two bodies be m each
• Then we can write: →FG=−[Gm2|→r|3]→r
• Where →r is the distance vector between A and B
2. Final mass of A = (m-0.25m) = 0.75m
• Final mass of B = (m+0.25m) = 1.25m
3. New force →FG(new)=−[G(0.75m)(1.25m)|→r|3]→r=−[0.9375Gm2|→r|3]→r
4. Dividing (3) by (1), we get:
→FG(new)→FG=(−[0.9375Gm2|→r|3]→r)÷(−[Gm2|→r|3]→r)=0.9375
⇒→FG(new)=0.9375→FG
Solved example 8.12
Two bodies A and B have masses m and M respectively. The gravitational force of attraction between them is →F. A third body C is placed in contact with A. The body C has a mass of 2m. What is the force on B due to A? What is the total force on B ?
Solution:
Part (a):
• Initially, there are only two masses A and B
• At that time, the force is given to be →F
• We can write:
→F=→FG(A,B)=−[GmM|→r|3]→r
• Where →r is the distance vector between A and B
Part (b):
• When C is placed near A, the force on B due to C is given by:
→FG(C,B)=−[G×2m×M|→r|3]→r
• Taking ratios, we get:
→FG(C,B)→FG(A,B)=→FG(C,B)→F=(−[G×2m×M|→r|3]→r)÷(−[G×m×M|→r|3]→r)
⇒→FG(C,B)→F=2
⇒→FG(C,B)=2→F
• Thus we can write:
Total force on B = →FG(A,B)+→FG(C,B)=→F+2→F=3→F
Solved example 8.13
In fig.8.15(a) below, three equal masses of m kg each are fixed at three corners A, B and D of the square of side a
![]() |
Fig.8.15 |
Solution:
1. Any square will lie on a plane. So this is a 2D problem. We will need the x and y axes only. There is no need for the z-axis
• Assume that, the origin of the coordinate axes is at A. This is shown in fig.b
2. For this square:
(i) C will be equidistant from B and D
C will be at a distance of (2)a from A
3. Now we can begin the calculations:
• The distance vectors are shown in fig.8.16(a) below:
![]() |
Fig.8.16 |
4. Force due to mass at B
(i) →FG(B,C)=−[Gm×1|→rB,C|3]→rB,C
⇒→FG(B,C)=−[Gma3]→rB,C
(ii) So our next task is to write →rB,C in component form
• To reach C from B:
♦ We have no horizontal distance to travel
♦ We travel only a distance BC vertically up
• So the components of →rB,C are (0)ˆi and aˆj
• We can write: →rB,C=aˆj
(iii) So the result in (i) becomes: →FG(B,C)=−[Gma3]aˆj
5. Force due to mass at D
(i) →FG(D,C)=−[Gm×1|→rD,C|3]→rD,C
⇒→FG(D,C)=−[Gma3]→rD,C
(ii) So our next task is to write →rD,C in component form
• To reach C from D:
♦ We have no vertical distance to travel
♦ We travel only a distance DC horizontally to the right
• So the components of →rD,C are (a)ˆi and 0ˆj
• We can write: →rD,C=aˆi
(iii) So the result in (i) becomes: →FG(D,C)=−[Gma3]aˆi
6. Force due to mass at A:
(i) →FG(A,C)=−[Gm×1|→rA,C|3]→rA,C
⇒→FG(A,C)=−[Gm(√2a)3]→rA,C
(ii) So our next task is to write →rA,C in component form
• To reach C from A:
♦ First we travel a distance AB horizontally to the right
♦ Then we travel a distance BC vertically up
• So the components of →rA,C are aˆi and aˆj
• We can write: →rA,C=aˆi+aˆj
(iii) So the result in (i) becomes:
→FG(A,C)=−[Gm(√2a)3](aˆi+aˆj)
⇒→FG(A,C)=(−[Gm(√2a)3]aˆi)+(−[Gm(√2a)3]aˆj)
7. When we add the results in (4), (5) and (6), we get the net force experienced by the mass at C
So we can write:
The net force experienced by the mass at C
= →FG(B,C)+→FG(D,C)+→FG(A,C)
= (−[Gma3]aˆj)+(−[Gma3]aˆi)+(−[Gm(√2a)3]aˆi)+(−[Gm(√2a)3]aˆj)
(i) First we will add the horizontal terms:
(−[Gma3]aˆi)+(−[Gm(√2a)3]aˆi)
= −[Gma3](aˆi+[12√2]aˆi)
= −[Gma2](1+12√2)ˆi
(ii) Next we add the vertical terms:
(−[Gma3]aˆj)+(−[Gm(√2a)3]aˆj)
= −[Gma3](aˆj+[12√2]aˆj)
= −[Gma2](1+12√2)ˆj
8. We can write:
■ The resultant force on C has two components:
• The horizontal component is: −[Gma2](1+12√2)ˆi
• The vertical component is: −[Gma2](1+12√2)ˆj
9. Note that, the magnitudes of both components are the same
So we can write:
■ The resultant force on C has two components:
• The horizontal component is: −pˆi
• The vertical component is: −pˆj
Where p=[Gma2](1+12√2)
This is shown in fig.8.17(a) below:
![]() |
Fig.8.17 |
♦ The magnitudes of the two vectors are equal
♦ The two vectors are perpendicular to each other
• So the magnitude of the resultant vector will be given by:
√p2+p2=√2p2=√2p (Details here)
= √2[Gma2](1+12√2)
= [Gma2](√2+12)
• Also, the resultant will be making an angle of 45o with the horizontal. This is shown in fig.8.17(b)
Another method:
• We can use trigonometry and write:
→rA,C=(acos45)ˆi+(asin45)ˆj
• The reader is advised to write all steps using this alternate method and prove that, the resultant force on the mass at C is the same as that obtained above
In the next section, we will see gravitational force of attraction between extended objects
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