
Concept explainers
a.
The mass of the Earth.
a.

Answer to Problem 41A
6.018×1024 kg
Explanation of Solution
Given:
The radius of Earth is 6.4×106 m .
Mass, m=1.0 kg
Weight, w=9.8 N
Formula used:
Gravitational force:
F=Gm1m2r2
Where, G is the gravitational constant, m1 and m2 are the masses, and r is the distance between the masses.
Calculation:
The value of gravitational constant is:
G=6.67×10−11 N.m2/kg2
Now, the gravitational attractive force exerted on the given mass by the Earth will be equal to the weight of the mass on Earth’s surface. Then using the given values of the mass and the radius of the Earth, the mass of the Earth will be,
F=GmEmr29.8=6.67×10−11×mE×1.0(6.4×106)2mE=6.018×1024 kg
Conclusion:
Thus, the mass of the Earth is 6.018×1024 kg .
b.
The average density of the Earth.
b.

Answer to Problem 41A
5.4×103 kg/m3
Explanation of Solution
Given:
The radius of Earth is 6.4×106 m .
From part (a), the mass of the Earth is 6.018×1024 kg .
Formula used:
Density = MassVolume
Calculation:
The volume of the Earth will be,
V=43πr3Earth
Then, the density of the Earth will be,
Density = (Mass)EarthVolume=6.018×102443π(6.4×106)3=5.4×103 kg/m3
Conclusion:
Thus, the average density of the Earth is 5.4×103 kg/m3 .
Chapter 7 Solutions
Glencoe Physics: Principles and Problems, Student Edition
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