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quester [9]
3 years ago
6

Find a numerical value for ρearth, the average density of the earth in kilograms per cubic meter. use 6378km for the radius of t

he earth, g=6.67×10−11m3/(kg⋅s2), and a value of g at the surface of 9.80m/s29.80m/s2.

Physics
2 answers:
Rufina [12.5K]3 years ago
5 0

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

\texttt{ }

<h3>Further explanation</h3>

Let's recall the Gravitational Force formula:

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

<em>where:</em>

<em>F = Gravitational Force ( N )</em>

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

<em>m = mass of object ( kg )</em>

<em>R = distance between object ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

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

<u>Asked:</u>

Average Density of the Earth = ρ = ?

<u>Solution:</u>

<em>Firstly , we will calculate the mass of the Earth as follows:</em>

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

\boxed {M = \frac{g R^2}{G} } → <em>Equation 1</em>

\texttt{ }

<em>Next , we could calculate the average density of the Earth as follows:</em>

\rho = M \div V

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

\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}

\texttt{ }

<h3>Learn more</h3>
  • 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

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Gravitational Force

Andre45 [30]3 years ago
4 0

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

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