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The earth has ratio 6400km the gravitational field strength on the earth surface is 9.81N/kg. Use these information determine a) mass of the earth b) It means tensity

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2 parts
a

6.02×1024 kg6.02 \times 10^{24} \text{ kg}

b

5480 kg/m35480 \text{ kg/m}^3

Step-by-step solution

6 steps

Here's the information provided:

  • Radius of the Earth, R=6400 km=6.4×106 mR = 6400 \text{ km} = 6.4 \times 10^6 \text{ m}
  • Gravitational field strength at the Earth's surface, g=9.81 N/kgg = 9.81 \text{ N/kg}
  • Gravitational constant, G=6.674×10−11 N m2/kg2G = 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2 (standard value)

Part a· 3 steps

Calculate the mass of the Earth

  1. The relationship between the gravitational field strength gg, the mass of the Earth MM, the radius of the Earth RR, and the gravitational constant GG is given by the formula

    g=GMR2g = \frac{GM}{R^2}

  2. To find the mass of the Earth MM, we rearrange the formula

    M=gR2GM = \frac{gR^2}{G}

  3. Substitute the numerical values into the formula

    M=9.81 N/kg×(6.4×106 m)26.674×10−11 N m2/kg2M = \frac{9.81 \text{ N/kg} \times (6.4 \times 10^6 \text{ m})^2}{6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2} M=9.81×(40.96×1012)6.674×10−11 kgM = \frac{9.81 \times (40.96 \times 10^{12})}{6.674 \times 10^{-11}} \text{ kg} M=401.6256×10126.674×10−11 kgM = \frac{401.6256 \times 10^{12}}{6.674 \times 10^{-11}} \text{ kg} M≈60.1776×1023 kgM \approx 60.1776 \times 10^{23} \text{ kg} M≈6.02×1024 kgM \approx 6.02 \times 10^{24} \text{ kg}

    The mass of the Earth is approximately 6.02×1024 kg\boxed{6.02 \times 10^{24} \text{ kg}}.

Part b

5480 kg/m35480 \text{ kg/m}^3

Calculate the mean density of the Earth

  1. The mean density ρ\rho is defined as the mass divided by the volume.

    For a sphere, the volume VV is given by V=43πR3V = \frac{4}{3}\pi R^3.

    ρ=MV=M43πR3\rho = \frac{M}{V} = \frac{M}{\frac{4}{3}\pi R^3}

  2. First, calculate the volume of the Earth using the radius R=6.4×106 mR = 6.4 \times 10^6 \text{ m}

    V=43π(6.4×106 m)3V = \frac{4}{3}\pi (6.4 \times 10^6 \text{ m})^3 V=43π(262.144×1018) m3V = \frac{4}{3}\pi (262.144 \times 10^{18}) \text{ m}^3 V≈1.098×1021 m3V \approx 1.098 \times 10^{21} \text{ m}^3

  3. Now, substitute the mass MM calculated previously and the volume VV into the density formula

    ρ=6.01776×1024 kg1.098066×1021 m3\rho = \frac{6.01776 \times 10^{24} \text{ kg}}{1.098066 \times 10^{21} \text{ m}^3} ρ≈5.480×103 kg/m3\rho \approx 5.480 \times 10^3 \text{ kg/m}^3 ρ≈5480 kg/m3\rho \approx 5480 \text{ kg/m}^3

    The mean density of the Earth is approximately 5480 kg/m3\boxed{5480 \text{ kg/m}^3}.

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