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Wednesday, January 8, 2020

Chapter 8.9 - Solved examples on Gravity

In the previous sectionwe 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(ms2)=6.67×1011(Nm2kg2)MM(1.74×106)2(m2)
1.67(ms2)=2.203×1023MM(Nkg2)
MM=7.58×1022(ms2)Nkg2
• We can put 'kg m s-2' instead of 'N'. So we get: 
MM=7.58×1022(ms2)kgms2kg2
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|=4×6.67×1011(Nm2kg2)×π×6.4×106(m)×11.34×103(kgm3)3
|g|=20.27(Nm2kg2)(m)(kgm3)
|g|=20.27(kgms2m2kg2)(m)(kgm3)
|g|=20.27ms2

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|(1dRE)
3. So we get: 0.5|g|=|g|(1d1RE)
0.5=(1d1RE)
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|(1dRE)
2. So we get: |g32|=|g|(1320006400000)
|g32|=|g|(11200)=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 |Δgh1|. At a particular depth d1, the acceleration due to gravity is lesser by |Δgd1|. If |Δgh1|=|Δgd1|, 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|(1dRE)
3. Given that |g||gh1|=|Δgh1|
• So we get: |Δgh1|=|g||g|(1+h1RE)2
4. Given that |g||gd1|=|Δgd1|
• So we get: |Δgd1|=|g||g|(1d1RE)
5. Given that: |Δgh1|=|Δgd1|
• So equating the results in (3) and (4), we get:
|g||g|(1+h1RE)2=|g||g|(1d1RE)
11(1+h1RE)2=1(1d1RE)
1(1+h1RE)2=(1d1RE)
6. Consider the left side of the above result. We have seen it's binomial expansion after discarding higher powers: 1(1+h1RE)2=(12h1RE)
• So the result in (5) becomes: (12h1RE)=(1d1RE)
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|(1dRE)
2. So we get: |g(0.5RE)|=|g|(10.5RERE)=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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