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: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
• 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: $\mathbf\small{|\vec{g}_M|=\frac{GM_M}{R_M^2}}$
Substituting the values, we get:
$\mathbf\small{1.67\;\rm{(m\;s^{-2})}=\frac{6.67\times 10^{-11}\rm{(N\;m^{2}\;kg^{-2})}M_M}{(1.74\times 10^{6})^2\rm{(m^2)}}}$
$\mathbf\small{\Rightarrow 1.67\;\rm{(m\;s^{-2})}=2.203\times 10^{-23}M_M\rm{(N\;kg^{-2})}}$
$\mathbf\small{\Rightarrow M_M=7.58 \times 10^{22}\frac{\rm{(m\;s^{-2})}}{\rm{N\,kg^{-2}}}}$
• We can put 'kg m s-2' instead of 'N'. So we get:
$\mathbf\small{M_M=7.58 \times 10^{22}\frac{\rm{(m\;s^{-2})}}{\rm{kg\,m\;s^{-2}\,kg^{-2}}}}$
$\mathbf\small{\Rightarrow M_M=7.58 \times 10^{22}\;\rm{kg}}$
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 $\mathbf\small{|\vec{g}|}$. What is the value of h1?
Solution:
1. Given that: $\mathbf\small{|\vec{g}_{h_1}|=\frac{|\vec{g}|}{64}}$
2. We have Eq.8.9 that we derived in the previous section: $\mathbf\small{|\vec{g}_h|=\frac{|\vec{g}|}{\left(1+\frac{h}{R_E}\right)^2}}$
3. So we get: $\mathbf\small{\frac{|\vec{g}|}{64}=\frac{|\vec{g}|}{\left(1+\frac{h_1}{R_E}\right)^2}}$
$\mathbf\small{\Rightarrow 64=\left(1+\frac{h_1}{R_E}\right)^2}$
$\mathbf\small{\Rightarrow 8=\left(1+\frac{h_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow \frac{h_1}{R_E}=7}$
$\mathbf\small{\Rightarrow h_1=7R_E}$ = 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 $\mathbf\small{|\vec{g}|}$?
Density of lead = 11.34 × 103 kg m-3
Solution:
1. We have Eq.8.7: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
2. If RE is the radius of the earth and $\mathbf\small{\rho}$ the density of lead, we have: $\mathbf\small{M_E=\frac{4\pi R_E^3 \rho}{3}}$
• Substituting this in (1), we get: $\mathbf\small{|\vec{g}|=\frac{G}{R_E^2}\times \frac{4\pi R_E^3 \rho}{3}}$
$\mathbf\small{\Rightarrow |\vec{g}|=\frac{4G\pi R_E \rho}{3}}$
3. Substituting the values, we get:
$\mathbf\small{|\vec{g}|=\frac{4\times 6.67\times 10^{-11}\rm{(N\;m^{2}\;kg^{-2})}\times \pi \times 6.4 \times 10^6 \rm{(m)} \times 11.34\times 10^{3}\rm{(kg\;m^{-3})}}{3}}$
$\mathbf\small{\Rightarrow |\vec{g}|=20.27\rm{(N\;m^{2}\;kg^{-2})}\rm{(m)}\rm{(kg\;m^{-3})}}$
$\mathbf\small{\Rightarrow |\vec{g}|=20.27\rm{(kg\,m\,s^{-2}\;m^{2}\;kg^{-2})}\rm{(m)}\rm{(kg\;m^{-3})}}$
$\mathbf\small{\Rightarrow |\vec{g}|=20.27\;\rm{m\,s^{-2}}}$
Solved example 8.18
On the surface of the earth, the acceleration due to gravity is $\mathbf\small{|\vec{g}|}$. At what depth d1 from the surface will the acceleration become $\mathbf\small{0.5|\vec{g}|}$?
Solution:
1. Given that: $\mathbf\small{|\vec{g}_{d_1}|=0.5|\vec{g}|}$
2. We have Eq.8.10 that we derived in the previous section: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
3. So we get: $\mathbf\small{0.5|\vec{g}|=|\vec{g}|\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow 0.5=\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow \frac{d_1}{R_E}=0.5}$
$\mathbf\small{\Rightarrow d_1=0.5R_E}$
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: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
2. So we get: $\mathbf\small{|\vec{g}_{32}|=|\vec{g}|\left(1- \frac{32000}{6400000}\right)}$
$\mathbf\small{\Rightarrow |\vec{g}_{32}|=|\vec{g}|\left(1- \frac{1}{200}\right)=0.995|\vec{g}|}$
3. Actual mass of the body = $\mathbf\small{\frac{m|\vec{g}|}{|\vec{g}|}=\frac{100}{|\vec{g}|}}$
4. So new weight = $\mathbf\small{\frac{100}{|\vec{g}|}\times 0.995|\vec{g}|=99.5\;N}$
Solved example 8.20
At a particular height h1, the acceleration due to gravity is lesser by $\mathbf\small{|\Delta \vec{g}_{h_1}|}$. At a particular depth d1, the acceleration due to gravity is lesser by $\mathbf\small{|\Delta \vec{g}_{d_1}|}$. If $\mathbf\small{|\Delta \vec{g}_{h_1}|=|\Delta \vec{g}_{d_1}|}$, what is the relation between d1 and h1?
Solution:
1. We have Eq.8.9 that we derived in the previous section: $\mathbf\small{|\vec{g}_h|=\frac{|\vec{g}|}{\left(1+\frac{h}{R_E}\right)^2}}$
2. We have Eq.8.10 that we derived in the previous section: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
3. Given that $\mathbf\small{|\vec{g}|-|\vec{g}_{h_1}|=|\Delta \vec{g}_{h_1}|}$
• So we get: $\mathbf\small{|\Delta \vec{g}_{h_1}|=|\vec{g}|-\frac{|\vec{g}|}{\left(1+\frac{h_1}{R_E}\right)^2}}$
4. Given that $\mathbf\small{|\vec{g}|-|\vec{g}_{d_1}|=|\Delta \vec{g}_{d_1}|}$
• So we get: $\mathbf\small{|\Delta \vec{g}_{d_1}|=|\vec{g}|-|\vec{g}|\left(1- \frac{d_1}{R_E}\right)}$
5. Given that: $\mathbf\small{|\Delta \vec{g}_{h_1}|=|\Delta \vec{g}_{d_1}|}$
• So equating the results in (3) and (4), we get:
$\mathbf\small{|\vec{g}|-\frac{|\vec{g}|}{\left(1+\frac{h_1}{R_E}\right)^2}=|\vec{g}|-|\vec{g}|\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow 1-\frac{1}{\left(1+\frac{h_1}{R_E}\right)^2}=1-\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow \frac{1}{\left(1+\frac{h_1}{R_E}\right)^2}=\left(1- \frac{d_1}{R_E}\right)}$
6. Consider the left side of the above result. We have seen it's binomial expansion after discarding higher powers: $\mathbf\small{\frac{1}{\left(1+\frac{h_1}{R_E}\right)^2}=\left(1- \frac{2h_1}{R_E}\right)}$
• So the result in (5) becomes: $\mathbf\small{\left(1- \frac{2h_1}{R_E}\right)=\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow \frac{2h_1}{R_E}=\frac{d_1}{R_E}}$
$\mathbf\small{\Rightarrow d_1 = 2h_1}$
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: $\mathbf\small{|\vec{g}_h|=\frac{|\vec{g}|}{\left(1+\frac{h}{R_E}\right)^2}}$
2. So we get: $\mathbf\small{|\vec{g}_{(0.5R_E)}|=\frac{|\vec{g}|}{\left(1+\frac{0.5R_E}{R_E}\right)^2}}$
$\mathbf\small{\Rightarrow |\vec{g}_{(0.5R_E)}|=\frac{|\vec{g}|}{\left(1+0.5\right)^2}=\frac{|\vec{g}|}{1.5^2}=\frac{|\vec{g}|}{2.25}}$
3. Actual mass of the body = $\mathbf\small{\frac{m|\vec{g}|}{|\vec{g}|}=\frac{63}{|\vec{g}|}}$
4. So new weight = $\mathbf\small{\frac{63}{|\vec{g}|}\times \frac{|\vec{g}|}{2.25}=28\, \rm{N}}$
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: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
2. So we get: $\mathbf\small{|\vec{g}_{(0.5R_E)}|=|\vec{g}|\left(1- \frac{0.5R_E}{R_E}\right)=0.5|\vec{g}|}$
3. Actual mass of the body = $\mathbf\small{\frac{m|\vec{g}|}{|\vec{g}|}=\frac{250}{|\vec{g}|}}$
4. So new weight = $\mathbf\small{\frac{250}{|\vec{g}|}\times 0.5|\vec{g}|=125\;\rm{N}}$
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: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
• 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: $\mathbf\small{|\vec{g}_M|=\frac{GM_M}{R_M^2}}$
Substituting the values, we get:
$\mathbf\small{1.67\;\rm{(m\;s^{-2})}=\frac{6.67\times 10^{-11}\rm{(N\;m^{2}\;kg^{-2})}M_M}{(1.74\times 10^{6})^2\rm{(m^2)}}}$
$\mathbf\small{\Rightarrow 1.67\;\rm{(m\;s^{-2})}=2.203\times 10^{-23}M_M\rm{(N\;kg^{-2})}}$
$\mathbf\small{\Rightarrow M_M=7.58 \times 10^{22}\frac{\rm{(m\;s^{-2})}}{\rm{N\,kg^{-2}}}}$
• We can put 'kg m s-2' instead of 'N'. So we get:
$\mathbf\small{M_M=7.58 \times 10^{22}\frac{\rm{(m\;s^{-2})}}{\rm{kg\,m\;s^{-2}\,kg^{-2}}}}$
$\mathbf\small{\Rightarrow M_M=7.58 \times 10^{22}\;\rm{kg}}$
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 $\mathbf\small{|\vec{g}|}$. What is the value of h1?
Solution:
1. Given that: $\mathbf\small{|\vec{g}_{h_1}|=\frac{|\vec{g}|}{64}}$
2. We have Eq.8.9 that we derived in the previous section: $\mathbf\small{|\vec{g}_h|=\frac{|\vec{g}|}{\left(1+\frac{h}{R_E}\right)^2}}$
3. So we get: $\mathbf\small{\frac{|\vec{g}|}{64}=\frac{|\vec{g}|}{\left(1+\frac{h_1}{R_E}\right)^2}}$
$\mathbf\small{\Rightarrow 64=\left(1+\frac{h_1}{R_E}\right)^2}$
$\mathbf\small{\Rightarrow 8=\left(1+\frac{h_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow \frac{h_1}{R_E}=7}$
$\mathbf\small{\Rightarrow h_1=7R_E}$ = 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 $\mathbf\small{|\vec{g}|}$?
Density of lead = 11.34 × 103 kg m-3
Solution:
1. We have Eq.8.7: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
2. If RE is the radius of the earth and $\mathbf\small{\rho}$ the density of lead, we have: $\mathbf\small{M_E=\frac{4\pi R_E^3 \rho}{3}}$
• Substituting this in (1), we get: $\mathbf\small{|\vec{g}|=\frac{G}{R_E^2}\times \frac{4\pi R_E^3 \rho}{3}}$
$\mathbf\small{\Rightarrow |\vec{g}|=\frac{4G\pi R_E \rho}{3}}$
3. Substituting the values, we get:
$\mathbf\small{|\vec{g}|=\frac{4\times 6.67\times 10^{-11}\rm{(N\;m^{2}\;kg^{-2})}\times \pi \times 6.4 \times 10^6 \rm{(m)} \times 11.34\times 10^{3}\rm{(kg\;m^{-3})}}{3}}$
$\mathbf\small{\Rightarrow |\vec{g}|=20.27\rm{(N\;m^{2}\;kg^{-2})}\rm{(m)}\rm{(kg\;m^{-3})}}$
$\mathbf\small{\Rightarrow |\vec{g}|=20.27\rm{(kg\,m\,s^{-2}\;m^{2}\;kg^{-2})}\rm{(m)}\rm{(kg\;m^{-3})}}$
$\mathbf\small{\Rightarrow |\vec{g}|=20.27\;\rm{m\,s^{-2}}}$
Solved example 8.18
On the surface of the earth, the acceleration due to gravity is $\mathbf\small{|\vec{g}|}$. At what depth d1 from the surface will the acceleration become $\mathbf\small{0.5|\vec{g}|}$?
Solution:
1. Given that: $\mathbf\small{|\vec{g}_{d_1}|=0.5|\vec{g}|}$
2. We have Eq.8.10 that we derived in the previous section: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
3. So we get: $\mathbf\small{0.5|\vec{g}|=|\vec{g}|\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow 0.5=\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow \frac{d_1}{R_E}=0.5}$
$\mathbf\small{\Rightarrow d_1=0.5R_E}$
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: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
2. So we get: $\mathbf\small{|\vec{g}_{32}|=|\vec{g}|\left(1- \frac{32000}{6400000}\right)}$
$\mathbf\small{\Rightarrow |\vec{g}_{32}|=|\vec{g}|\left(1- \frac{1}{200}\right)=0.995|\vec{g}|}$
3. Actual mass of the body = $\mathbf\small{\frac{m|\vec{g}|}{|\vec{g}|}=\frac{100}{|\vec{g}|}}$
4. So new weight = $\mathbf\small{\frac{100}{|\vec{g}|}\times 0.995|\vec{g}|=99.5\;N}$
Solved example 8.20
At a particular height h1, the acceleration due to gravity is lesser by $\mathbf\small{|\Delta \vec{g}_{h_1}|}$. At a particular depth d1, the acceleration due to gravity is lesser by $\mathbf\small{|\Delta \vec{g}_{d_1}|}$. If $\mathbf\small{|\Delta \vec{g}_{h_1}|=|\Delta \vec{g}_{d_1}|}$, what is the relation between d1 and h1?
Solution:
1. We have Eq.8.9 that we derived in the previous section: $\mathbf\small{|\vec{g}_h|=\frac{|\vec{g}|}{\left(1+\frac{h}{R_E}\right)^2}}$
2. We have Eq.8.10 that we derived in the previous section: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
3. Given that $\mathbf\small{|\vec{g}|-|\vec{g}_{h_1}|=|\Delta \vec{g}_{h_1}|}$
• So we get: $\mathbf\small{|\Delta \vec{g}_{h_1}|=|\vec{g}|-\frac{|\vec{g}|}{\left(1+\frac{h_1}{R_E}\right)^2}}$
4. Given that $\mathbf\small{|\vec{g}|-|\vec{g}_{d_1}|=|\Delta \vec{g}_{d_1}|}$
• So we get: $\mathbf\small{|\Delta \vec{g}_{d_1}|=|\vec{g}|-|\vec{g}|\left(1- \frac{d_1}{R_E}\right)}$
5. Given that: $\mathbf\small{|\Delta \vec{g}_{h_1}|=|\Delta \vec{g}_{d_1}|}$
• So equating the results in (3) and (4), we get:
$\mathbf\small{|\vec{g}|-\frac{|\vec{g}|}{\left(1+\frac{h_1}{R_E}\right)^2}=|\vec{g}|-|\vec{g}|\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow 1-\frac{1}{\left(1+\frac{h_1}{R_E}\right)^2}=1-\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow \frac{1}{\left(1+\frac{h_1}{R_E}\right)^2}=\left(1- \frac{d_1}{R_E}\right)}$
6. Consider the left side of the above result. We have seen it's binomial expansion after discarding higher powers: $\mathbf\small{\frac{1}{\left(1+\frac{h_1}{R_E}\right)^2}=\left(1- \frac{2h_1}{R_E}\right)}$
• So the result in (5) becomes: $\mathbf\small{\left(1- \frac{2h_1}{R_E}\right)=\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow \frac{2h_1}{R_E}=\frac{d_1}{R_E}}$
$\mathbf\small{\Rightarrow d_1 = 2h_1}$
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: $\mathbf\small{|\vec{g}_h|=\frac{|\vec{g}|}{\left(1+\frac{h}{R_E}\right)^2}}$
2. So we get: $\mathbf\small{|\vec{g}_{(0.5R_E)}|=\frac{|\vec{g}|}{\left(1+\frac{0.5R_E}{R_E}\right)^2}}$
$\mathbf\small{\Rightarrow |\vec{g}_{(0.5R_E)}|=\frac{|\vec{g}|}{\left(1+0.5\right)^2}=\frac{|\vec{g}|}{1.5^2}=\frac{|\vec{g}|}{2.25}}$
3. Actual mass of the body = $\mathbf\small{\frac{m|\vec{g}|}{|\vec{g}|}=\frac{63}{|\vec{g}|}}$
4. So new weight = $\mathbf\small{\frac{63}{|\vec{g}|}\times \frac{|\vec{g}|}{2.25}=28\, \rm{N}}$
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: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
2. So we get: $\mathbf\small{|\vec{g}_{(0.5R_E)}|=|\vec{g}|\left(1- \frac{0.5R_E}{R_E}\right)=0.5|\vec{g}|}$
3. Actual mass of the body = $\mathbf\small{\frac{m|\vec{g}|}{|\vec{g}|}=\frac{250}{|\vec{g}|}}$
4. So new weight = $\mathbf\small{\frac{250}{|\vec{g}|}\times 0.5|\vec{g}|=125\;\rm{N}}$
• In the next section, we will see gravitational field and it's intensity
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