Using Newton's law of gravitation, we can determine the mass of the Earth and its mean density. These quantities are calculated using the values of acceleration due to gravity, radius of the Earth, and the universal gravitational constant.
Mass of the Earth
Consider a body of mass \(m\) placed on the surface of the Earth.
According to Newton's law of gravitation, the gravitational force acting on the body is:
\[ F=\frac{GMm}{R^2} \]
Where:
- \(M\) = Mass of the Earth
- \(R\) = Radius of the Earth
- \(G\) = Universal Gravitational Constant
- \(m\) = Mass of the body
The weight of the body is:
\[ F=mg \]
Equating the two expressions:
\[ mg=\frac{GMm}{R^2} \]
Cancelling \(m\) from both sides:
\[ g=\frac{GM}{R^2} \]
Therefore,
\[ M=\frac{gR^2}{G} \]
Calculation of Earth's Mass
Given:
\[ g=9.8\ \text{m s}^{-2} \]
\[ R=6400\ \text{km}=6.4\times10^6\ \text{m} \]
\[ G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2} \]
Substituting the values:
\[ M=\frac{9.8(6.4\times10^6)^2}{6.67\times10^{-11}} \]
\[ M=5.98\times10^{24}\ \text{kg} \]
Result
\[ \boxed{M=5.98\times10^{24}\ \text{kg}} \]
Thus, the mass of the Earth is approximately \(5.98\times10^{24}\) kg.
Mean Density of Earth
Let \(\rho\) be the mean density of the Earth.
We know:
\[ g=\frac{GM}{R^2} \]
Assuming Earth is a homogeneous sphere:
\[ M=\text{Volume}\times\text{Density} \]
\[ M=\frac{4}{3}\pi R^3\rho \]
Substituting this value into the equation for \(g\):
\[ g=\frac{G}{R^2}\left(\frac{4}{3}\pi R^3\rho\right) \]
\[ g=\frac{4}{3}\pi GR\rho \]
Therefore,
\[ \rho=\frac{3g}{4\pi GR} \]
Calculation of Mean Density
\[ g=9.8\ \text{m s}^{-2} \]
\[ G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2} \]
\[ R=6.4\times10^6\ \text{m} \]
\[ \rho= \frac{3\times9.8} {4\times3.142\times6.67\times10^{-11}\times6.4\times10^6} \]
\[ \rho=5478.4\ \text{kg m}^{-3} \]
Result
\[ \boxed{\rho=5478.4\ \text{kg m}^{-3}} \]
Thus, the mean density of the Earth is \(5478.4\ \text{kg m}^{-3}\).
Important Fact
Density of water:
\[ 1000\ \text{kg m}^{-3} \]
Ratio of Earth's density to water's density:
\[ \frac{5478.4}{1000}\approx5.5 \]
Therefore, the Earth is approximately 5.5 times denser than water.
Frequently Asked Questions (FAQs)
1. What is the mass of the Earth?
\[ 5.98\times10^{24}\ \text{kg} \]
2. Which law is used to calculate the mass of the Earth?
Newton's Law of Gravitation.
3. What is the value of the universal gravitational constant?
\[ G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2} \]
4. What is the mean density of the Earth?
\[ 5478.4\ \text{kg m}^{-3} \]
5. What is the formula for the mean density of Earth?
\[ \rho=\frac{3g}{4\pi GR} \]
6. What is the radius of Earth used in the calculation?
\[ R=6.4\times10^6\ \text{m} \]
7. How many times is Earth denser than water?
Approximately 5.5 times.
8. What is the SI unit of density?
\[ \text{kg m}^{-3} \]
Quiz: Mass of the Earth and Mean Density of Earth
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Which law is used to calculate the mass of the Earth?
A) Ohm's Law
B) Newton's Law of Gravitation
C) Boyle's Law
D) Hooke's LawAnswer: B) Newton's Law of Gravitation
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The gravitational force acting on a body of mass \(m\) is:
A) \(F=mgR\)
B) \(F=\frac{GMm}{R^2}\)
C) \(F=GMmR^2\)
D) \(F=\frac{GM}{R}\)Answer: B
-
The formula for the mass of the Earth is:
A) \(M=\frac{gR}{G}\)
B) \(M=\frac{gR^2}{G}\)
C) \(M=\frac{GR^2}{g}\)
D) \(M=\frac{g^2R}{G}\)Answer: B
-
The mass of the Earth is:
A) \(5.98\times10^{20}\) kg
B) \(5.98\times10^{22}\) kg
C) \(5.98\times10^{24}\) kg
D) \(5.98\times10^{26}\) kgAnswer: C
-
The value of acceleration due to gravity is:
A) \(10.8\ \text{m s}^{-2}\)
B) \(9.8\ \text{m s}^{-2}\)
C) \(8.9\ \text{m s}^{-2}\)
D) \(9.0\ \text{m s}^{-2}\)Answer: B
-
The formula for mean density of Earth is:
A) \(\rho=\frac{4\pi GR}{3g}\)
B) \(\rho=\frac{3g}{4\pi GR}\)
C) \(\rho=\frac{g}{GR}\)
D) \(\rho=\frac{GR}{g}\)Answer: B
-
The mean density of Earth is:
A) \(547.84\ \text{kg m}^{-3}\)
B) \(5478.4\ \text{kg m}^{-3}\)
C) \(54784\ \text{kg m}^{-3}\)
D) \(54.784\ \text{kg m}^{-3}\)Answer: B
-
Earth is approximately how many times denser than water?
A) 2.5
B) 3.5
C) 4.5
D) 5.5Answer: D
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