Question

Find a numerical value for ρearth, the average density of the earth in kilograms per cubic meter. use 6378km for the radius of the earth, g=6.67×10−11m3/(kg⋅s2), and a value of g at the surface of 9.80m/s29.80m/s2.

Answer 1

The average density of the earth is about 5.50 × 10³ kg/m³

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Further explanation

Let's recall the Gravitational Force formula:

$\boxed {F = G\ \frac{m_1 m_2}{R^2}}$

where:

F = Gravitational Force ( N )

G = Gravitational Constant ( = 6.67 × 10⁻¹¹ Nm²/kg² )

m = mass of object ( kg )

R = distance between object ( m )

Let us now tackle the problem!

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Given:

Radius of The Earth = R = 6378 km = 6378000 m

Gravitational Constant = G = 6.67 × 10⁻¹¹ Nm²/kg²

Gravitational Acceleration = g = 9.90 m/s²

Asked:

Average Density of the Earth = ρ = ?

Solution:

Firstly , we will calculate the mass of the Earth as follows:

$g = G \frac{M}{R^2}$

$\boxed {M = \frac{g R^2}{G} }$ → Equation 1

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Next , we could calculate the average density of the Earth as follows:

$\rho = M \div V$

$\rho = (\frac{g R^2}{G}) \div ( \frac{4}{3} \pi R^3 )$  ← Equation 1

$\rho = \frac{3gR^2}{4 \pi G R^3}$

$\rho = \frac{3g}{4 \pi G R}$

$\rho = \frac{3 \times 9.80}{ 4 \pi \times 6.67 \times 10^{-11} \times 6378000 }$

$\boxed {\rho \approx 5.50 \times 10^3 \texttt{ kg/m}^3}$

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Learn more

• Unit of G : brainly.com/question/1724648
• Velocity of Runner : brainly.com/question/3813437

• Kinetic Energy : brainly.com/question/692781

• Acceleration : brainly.com/question/2283922

• The Speed of Car : brainly.com/question/568302

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Answer details

Grade: High School

Subject: Mathematics

Chapter: Gravitational Force

Answer 2

Answer:

5501 kg/m^3

Explanation:

The value of g at the Earth's surface is

$g=\frac{GM}{R^2}=9.70 m/s^2$

where G is the gravitational constant

M is the Earth's mass

$R=6378km = 6.378 \cdot 10^6 m$ is the Earth's radius

Solving the formula for M, we find the value of the Earth's mass:

$M=\frac{gR^2}{G}=\frac{(9.81 m/s^2)(6.378\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=5.98\cdot 10^{24}kg$

The Earth's volume is (approximating the Earth to a perfect sphere)

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6.378\cdot 10^6 m)^3=1.087\cdot 10^{21} m^3$

So, the average density of the Earth is

$\rho = \frac{M}{V}=\frac{5.98\cdot 10^{24} kg}{1.087\cdot 10^{21} m^3}=5501 kg/m^3$