Question

An Earthlike Planet- In January 2006 astronomers reported the discovery of a planet, comparable in size to the earth,orbiting another star and having a mass about 5.5 times the earth’s mass. It is believed to consist of a mixture of rock andice, similar to Neptune. If this planet has the same density as Neptune (1.76 g/cm3), what is its radius expressed (a) inkilometers and (b) as a multiple of earth’s radius? Consult Appendix F for astronomical data.

Answer 1

To solve this problem we will begin by applying the given relations of density in terms of mass and volume, and from this last value we will take its geometric measurement for a sphere (Approximation of a planet) From there we will find the radius of the planet. Finally we will make a comparison between the radius of the new planet and the radius of the earth to understand its proportion.

Defining the Volume variables we have to

$V = \frac{m}{\rho}$

Here

V= Volume

m = mass

$\rho$=Density

For a spherical object the Volume is

$V = \frac{4}{3} \pi R^3$

PART A)

Equation we have

$\frac{4}{3} \pi r^3 = \frac{m}{\rho}$

$R = (\frac{3}{4\pi}(\frac{m}{\rho}))^{1/3}$

In this case the mass of new planet is 5.5times the mass of Earth,

$m = 5.5m_E$

Then,

$R = (\frac{3}{4\pi}(\frac{5.5m_E}{\rho}))^{1/3}$

The mass of the Earth is $5.97*10^{24}$ kg and the density is $1.76g/cm^3 = 1760 kg/m^3$,

Replacing we have that,

$R = (\frac{3}{4\pi}(\frac{5.5(5.97*10^{24})}{1.76}))^{1/3}$

$R = 1.64*10^7m$

$R = 1.64*10^4 Km$

Therefore the radius of this new planet is $R = 1.64*10^4 Km$

PART B) The value of radius of the Earth is

$R_E = 6.37*10^3 km$

Then the relation between them is

$\frac{R}{R_E}= \frac{1.64*10^4}{6.37*10^3}$

$\frac{R}{R_E}= 2.57$

Therefore the radius of the new planet in terms of radius of the Earth is $2.57R_E$