In the previous section, we completed a discussion on the basics of gravitational force, gravity and acceleration due to gravity. We saw a solved example also. In this section, we will see a few more solved examples
Solved example 8.15
The acceleration due to gravitational force at the surface of the moon is 1.67 ms-2. If the radius of the moon is 1.74 × 106 m, calculate the mass of the moon
Solution:
1. For earth, we have: |→g|=GMER2E
• We derived this formula by assuming that the earth is made up of concentric shells
• This assumption can be applied to any planet or heavenly body having a spherical shape
2. So for the moon, we can write: |→gM|=GMMR2M
Substituting the values, we get:
1.67(ms−2)=6.67×10−11(Nm2kg−2)MM(1.74×106)2(m2)
⇒1.67(ms−2)=2.203×10−23MM(Nkg−2)
⇒MM=7.58×1022(ms−2)Nkg−2
• We can put 'kg m s-2' instead of 'N'. So we get:
MM=7.58×1022(ms−2)kgms−2kg−2
⇒MM=7.58×1022kg
Solved example 8.16
At a particular height h1 above the surface of the earth, the value of acceleration due to gravity is 1/64 of |→g|. What is the value of h1?
Solution:
1. Given that: |→gh1|=|→g|64
2. We have Eq.8.9 that we derived in the previous section: |→gh|=|→g|(1+hRE)2
3. So we get: |→g|64=|→g|(1+h1RE)2
⇒64=(1+h1RE)2
⇒8=(1+h1RE)
⇒h1RE=7
⇒h1=7RE = 7 × 6.378 × 106 m = 44.65 × 106 m
Solved example 8.17
Assume that, earth is made up of lead. What would be the value of |→g|?
Density of lead = 11.34 × 103 kg m-3
Solution:
1. We have Eq.8.7: |→g|=GMER2E
2. If RE is the radius of the earth and ρ the density of lead, we have: ME=4πR3Eρ3
• Substituting this in (1), we get: |→g|=GR2E×4πR3Eρ3
⇒|→g|=4GπREρ3
3. Substituting the values, we get:
|→g|=4×6.67×10−11(Nm2kg−2)×π×6.4×106(m)×11.34×103(kgm−3)3
⇒|→g|=20.27(Nm2kg−2)(m)(kgm−3)
⇒|→g|=20.27(kgms−2m2kg−2)(m)(kgm−3)
⇒|→g|=20.27ms−2
Solved example 8.18
On the surface of the earth, the acceleration due to gravity is |→g|. At what depth d1 from the surface will the acceleration become 0.5|→g|?
Solution:
1. Given that: |→gd1|=0.5|→g|
2. We have Eq.8.10 that we derived in the previous section: |→gd|=|→g|(1−dRE)
3. So we get: 0.5|→g|=|→g|(1−d1RE)
⇒0.5=(1−d1RE)
⇒d1RE=0.5
⇒d1=0.5RE
Solved example 8.19
A body weighs 100 N on the surface of the earth. It is taken down to a depth of 32 km below the surface. What will be it's new weight? Take radius of the earth as 6400 km
Solution:
1. We have Eq.8.10 that we derived in the previous section: |→gd|=|→g|(1−dRE)
2. So we get: |→g32|=|→g|(1−320006400000)
⇒|→g32|=|→g|(1−1200)=0.995|→g|
3. Actual mass of the body = m|→g||→g|=100|→g|
4. So new weight = 100|→g|×0.995|→g|=99.5N
Solved example 8.20
At a particular height h1, the acceleration due to gravity is lesser by |Δ→gh1|. At a particular depth d1, the acceleration due to gravity is lesser by |Δ→gd1|. If |Δ→gh1|=|Δ→gd1|, what is the relation between d1 and h1?
Solution:
1. We have Eq.8.9 that we derived in the previous section: |→gh|=|→g|(1+hRE)2
2. We have Eq.8.10 that we derived in the previous section: |→gd|=|→g|(1−dRE)
3. Given that |→g|−|→gh1|=|Δ→gh1|
• So we get: |Δ→gh1|=|→g|−|→g|(1+h1RE)2
4. Given that |→g|−|→gd1|=|Δ→gd1|
• So we get: |Δ→gd1|=|→g|−|→g|(1−d1RE)
5. Given that: |Δ→gh1|=|Δ→gd1|
• So equating the results in (3) and (4), we get:
|→g|−|→g|(1+h1RE)2=|→g|−|→g|(1−d1RE)
⇒1−1(1+h1RE)2=1−(1−d1RE)
⇒1(1+h1RE)2=(1−d1RE)
6. Consider the left side of the above result. We have seen it's binomial expansion after discarding higher powers: 1(1+h1RE)2=(1−2h1RE)
• So the result in (5) becomes: (1−2h1RE)=(1−d1RE)
⇒2h1RE=d1RE
⇒d1=2h1
Solved example 8.21
A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth?
Solution:
1. We have Eq.8.9 that we derived in the previous section: |→gh|=|→g|(1+hRE)2
2. So we get: |→g(0.5RE)|=|→g|(1+0.5RERE)2
⇒|→g(0.5RE)|=|→g|(1+0.5)2=|→g|1.52=|→g|2.25
3. Actual mass of the body = m|→g||→g|=63|→g|
4. So new weight = 63|→g|×|→g|2.25=28N
Solved example 2.22
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the center of the earth, if it weighed 250 N on the surface?
Solution:
1. We have Eq.8.10 that we derived in the previous section: |→gd|=|→g|(1−dRE)
2. So we get: |→g(0.5RE)|=|→g|(1−0.5RERE)=0.5|→g|
3. Actual mass of the body = m|→g||→g|=250|→g|
4. So new weight = 250|→g|×0.5|→g|=125N
Solved example 8.15
The acceleration due to gravitational force at the surface of the moon is 1.67 ms-2. If the radius of the moon is 1.74 × 106 m, calculate the mass of the moon
Solution:
1. For earth, we have: |→g|=GMER2E
• We derived this formula by assuming that the earth is made up of concentric shells
• This assumption can be applied to any planet or heavenly body having a spherical shape
2. So for the moon, we can write: |→gM|=GMMR2M
Substituting the values, we get:
1.67(ms−2)=6.67×10−11(Nm2kg−2)MM(1.74×106)2(m2)
⇒1.67(ms−2)=2.203×10−23MM(Nkg−2)
⇒MM=7.58×1022(ms−2)Nkg−2
• We can put 'kg m s-2' instead of 'N'. So we get:
MM=7.58×1022(ms−2)kgms−2kg−2
⇒MM=7.58×1022kg
Solved example 8.16
At a particular height h1 above the surface of the earth, the value of acceleration due to gravity is 1/64 of |→g|. What is the value of h1?
Solution:
1. Given that: |→gh1|=|→g|64
2. We have Eq.8.9 that we derived in the previous section: |→gh|=|→g|(1+hRE)2
3. So we get: |→g|64=|→g|(1+h1RE)2
⇒64=(1+h1RE)2
⇒8=(1+h1RE)
⇒h1RE=7
⇒h1=7RE = 7 × 6.378 × 106 m = 44.65 × 106 m
Solved example 8.17
Assume that, earth is made up of lead. What would be the value of |→g|?
Density of lead = 11.34 × 103 kg m-3
Solution:
1. We have Eq.8.7: |→g|=GMER2E
2. If RE is the radius of the earth and ρ the density of lead, we have: ME=4πR3Eρ3
• Substituting this in (1), we get: |→g|=GR2E×4πR3Eρ3
⇒|→g|=4GπREρ3
3. Substituting the values, we get:
|→g|=4×6.67×10−11(Nm2kg−2)×π×6.4×106(m)×11.34×103(kgm−3)3
⇒|→g|=20.27(Nm2kg−2)(m)(kgm−3)
⇒|→g|=20.27(kgms−2m2kg−2)(m)(kgm−3)
⇒|→g|=20.27ms−2
Solved example 8.18
On the surface of the earth, the acceleration due to gravity is |→g|. At what depth d1 from the surface will the acceleration become 0.5|→g|?
Solution:
1. Given that: |→gd1|=0.5|→g|
2. We have Eq.8.10 that we derived in the previous section: |→gd|=|→g|(1−dRE)
3. So we get: 0.5|→g|=|→g|(1−d1RE)
⇒0.5=(1−d1RE)
⇒d1RE=0.5
⇒d1=0.5RE
Solved example 8.19
A body weighs 100 N on the surface of the earth. It is taken down to a depth of 32 km below the surface. What will be it's new weight? Take radius of the earth as 6400 km
Solution:
1. We have Eq.8.10 that we derived in the previous section: |→gd|=|→g|(1−dRE)
2. So we get: |→g32|=|→g|(1−320006400000)
⇒|→g32|=|→g|(1−1200)=0.995|→g|
3. Actual mass of the body = m|→g||→g|=100|→g|
4. So new weight = 100|→g|×0.995|→g|=99.5N
Solved example 8.20
At a particular height h1, the acceleration due to gravity is lesser by |Δ→gh1|. At a particular depth d1, the acceleration due to gravity is lesser by |Δ→gd1|. If |Δ→gh1|=|Δ→gd1|, what is the relation between d1 and h1?
Solution:
1. We have Eq.8.9 that we derived in the previous section: |→gh|=|→g|(1+hRE)2
2. We have Eq.8.10 that we derived in the previous section: |→gd|=|→g|(1−dRE)
3. Given that |→g|−|→gh1|=|Δ→gh1|
• So we get: |Δ→gh1|=|→g|−|→g|(1+h1RE)2
4. Given that |→g|−|→gd1|=|Δ→gd1|
• So we get: |Δ→gd1|=|→g|−|→g|(1−d1RE)
5. Given that: |Δ→gh1|=|Δ→gd1|
• So equating the results in (3) and (4), we get:
|→g|−|→g|(1+h1RE)2=|→g|−|→g|(1−d1RE)
⇒1−1(1+h1RE)2=1−(1−d1RE)
⇒1(1+h1RE)2=(1−d1RE)
6. Consider the left side of the above result. We have seen it's binomial expansion after discarding higher powers: 1(1+h1RE)2=(1−2h1RE)
• So the result in (5) becomes: (1−2h1RE)=(1−d1RE)
⇒2h1RE=d1RE
⇒d1=2h1
Solved example 8.21
A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth?
Solution:
1. We have Eq.8.9 that we derived in the previous section: |→gh|=|→g|(1+hRE)2
2. So we get: |→g(0.5RE)|=|→g|(1+0.5RERE)2
⇒|→g(0.5RE)|=|→g|(1+0.5)2=|→g|1.52=|→g|2.25
3. Actual mass of the body = m|→g||→g|=63|→g|
4. So new weight = 63|→g|×|→g|2.25=28N
Solved example 2.22
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the center of the earth, if it weighed 250 N on the surface?
Solution:
1. We have Eq.8.10 that we derived in the previous section: |→gd|=|→g|(1−dRE)
2. So we get: |→g(0.5RE)|=|→g|(1−0.5RERE)=0.5|→g|
3. Actual mass of the body = m|→g||→g|=250|→g|
4. So new weight = 250|→g|×0.5|→g|=125N
• In the next section, we will see gravitational field and it's intensity
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