Weighing the Earth by Its Pull

solved UniversalGravitation

Q. Weighing the Earth: you are given g = 9.81 m/s² $g_{s} = 9.81$ m/s² at the Earth's surface and the Earth's radius 6.37 × 10⁶ m $R = 6.37 \times 10^{6}$ m. What is the mass of the Earth, in kg?

Inputs

gravity

g_{s}

9.81 m/s²

radius

R

6370000.00 m

Outputs

mass

m

How it's solved

Solved with UniversalGravitation

Given

$g_{s} = 9.81$ m/s²
$R = 6.37 \times 10^{6}$ m

Find $m$ .

Working

gravity at the surface(earth) $g_{s} = \frac{G \cdot m}{R^{2}}$
$9.81 = \frac{6.674 \times 10^{−11} \cdot m}{{(6.37 \times 10^{6})}^{2}}$
$m = 5.964 \times 10^{24}$ kg
$m \approx 5.96 \times 10^{24}$ kgto 3 significant figures

Conclusion

How hard the Earth pulls at its surface depends only on its mass and how far the surface is from its centre, so the pull and the radius give the mass.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :weighing_the_earth_by_its_pull do4  kind :numerical5  describes "Weighing the Earth: you are given g = %{gravity} at the Earth's surface and the Earth's radius " \6            "%{radius}. %{mass}, in kg?"7  given gravity: "9.81 m/s2"8  given radius: "6.37e6 m"9  asks mass: "What is the mass of the Earth"10end

solution.rb

1require_relative "../questions/weighing_the_earth_by_its_pull"2require_relative "../science/gravitation/universal_gravitation"34solve :weighing_the_earth_by_its_pull, using: UniversalGravitation do5  objects :earth67  take gravity: earth.surface_gravity8  take radius: earth.radius_of_the_body910  mass = earth.mass11  answer mass:, in: "kg", figures: 31213  conclusion "How hard the Earth pulls at its surface depends only on its mass and how far the surface " \14             "is from its centre, so the pull and the radius give the mass."15end

playground.rb

1require_relative "../solutions/weighing_the_earth_by_its_pull"23play_with :weighing_the_earth_by_its_pull do4  title "Weighing the Earth by Its Pull"5end

The laws it drew on

quantity	written	units	within	from
gravitational constant	$G$	N m² kg⁻²	—	UniversalGravitation
mass	$m$	kg	0 … ∞	UniversalGravitation
radius of the body	$R$	m	0 … ∞	UniversalGravitation
surface gravity	$g_{s}$	m/s²	—	UniversalGravitation
separation	$r_{g}$	m	0 … ∞	UniversalGravitation
gravitational intensity	$E_{g}$	m/s²	—	UniversalGravitation
surface gravity of ground	$g_{s,g}$	m/s²	—
mass of ground	$m_{g}$	kg	0 … ∞
radius of the body of ground	$R_{g}$	m	0 … ∞
surface gravity of earth	$g_{s}$	m/s²	—
mass of earth	$m$	kg	0 … ∞
radius of the body of earth	$R$	m	0 … ∞
gravitational intensity of ground by earth	$E_{g,g}$	m/s²	—
separation of ground from earth	$r_{g,g}$	m	0 … ∞
gravitational intensity of earth by ground	$E_{g,g}$	m/s²	—
separation of earth from ground	$r_{g}$	m	0 … ∞

g_{s,b} = \frac{G \cdot m_{b}}{{R_{b}}^{2}}

gravity at the surface

E_{g,oo} = \frac{G \cdot m_{o}}{{r_{g,oo}}^{2}}

law of gravitation

g_{s,g} = \frac{G \cdot m_{g}}{{R_{g}}^{2}}

gravity at the surface(ground)

g_{s} = \frac{G \cdot m}{R^{2}}

gravity at the surface(earth)

E_{g,g} = \frac{G \cdot m}{{r_{g,g}}^{2}}

law of gravitation(ground, earth)

E_{g,g} = \frac{G \cdot m_{g}}{{r_{g}}^{2}}

law of gravitation(earth, ground)