Question

G the density of an object equals its mass divided by its volume. the mass of earth is 6 × 1024 kg and its radius is 4 × 103 miles. the mass of the sun is 2 × 1033 g and its radius is 7 × 105 km. calculate the earth's density divided by that of the sun.

Answer 1

Assuming the earth and the sun to be perfect spheres,

Volume of the sphere = 4/3 * pi * (r**3)

Volume of the earth = 4/3 * pi * ((4000*1.609 km)**3) = 1.116 E 12 km3

Volume of the Sun= 4/3 * pi * ((7 E 5 km)**3) = 1.436 E 18 km3

Density = mass /volume

Density of earth = 6 E 24 kg / 1.116 E 12 km3 = 5.376 E 12 [kg/km3]

Density of Sun= 2 E 30  kg / 1.436 E 18 km3 = 1.392 E 12 [kg/km3]


Density of earth / Density of Sun =

5.376 E 12 [kg/km3] / 1.392 E 12 [kg/km3] = 3.86

Answer 2

The earth's density divided by that of the sun = 3.85.10⁻³

Density is the ratio of mass per unit volume

Further explanation

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than the low density

The unit of density can be expressed in $\frac{g}{cm^3}\:or\:\frac{kg}{m^3}$

Density formula:

$\large{\boxed{\bold{\rho\:=\:\frac{m}{V} }}}$

ρ = density

m = mass

v = volume

We consider the Earth and the Sun to be spheres

We can know from the task that :

m earth = 6.10²⁴ kg

r earth = 4.10³ miles = 6,436.10⁶ m

earth volume =

$\frac{4}{3} \pi r^3$

$=\:\frac{4}{3} \pi(6.436.10^6)^3$

V earth = 1.12.10²¹ m³

m sun = 2.10³³ kg

r sun = 7.10⁵ miles = 7.10⁸ m

sun volume =

$\frac{4}{3} \pi r^3$

$=\:\frac{4}{3} \pi( 7.10^8)^3$

V sun = 1.436 .10²⁷m³

So

$\rho_{earth}\:=\:\frac{6.10^{24}}{1.12.10^{21}}$

$\rho_{earth}\:=\:5.36.10^3\:\frac{kg}{m^3}$

$\rho_{sun}\:=\:\frac{2.10^{33}}{1.436^{27}}$

$\rho_{sun}\:=\:1.393.10^6\:\frac{kg}{m^3}$

Then :

$\frac{\rho_{earth}}{\rho_{sun}}\:=\:\frac{5.36.10^3}{1.393.10^6}$

ρ earth : ρ sun ratio = 3.85.10⁻³

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Keywords: density's sun, density's earth, mass, volume, spheres,miles, km