Q51P - University Physics with Modern Physics

Question

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 and ice, similar to Neptune. If this planet has the same density as Neptune $1.76 \mathrm{g}/{\mathrm{cm}}^3$, what is its radius expressed (a) in kilometres and (b) as a multiple of earth’s radius? Consult Appendix F for astronomical data.

Step-by-Step Solution

Answer

1. The planet’s radius is $1.64\times {10}^4 \mathrm{km}$.
2. The planet’s radius in terms of the earth’s radius $\left(r_E\right)$ is $2.57 r_E$.

Significance of Density and volume

The volume of an object is the amount of three-dimensional space it takes up, whereas density is the mass of the unit volume of the object.

Identification of given data

The given data and data from Appendix $F$ can be listed below as,

The mass of the planet is,

$\begin{array}{rcl}{M}_P& =& 5.5M_E\\ & =& (5.5)\left(5.97\times {10}^{24} \mathrm{kg}\right).\\ & =& 3.28\times {10}^{25} \mathrm{kg}\end{array}$

The density of the planet is, ${\rho }_P=1.76 \mathrm{g}/{\mathrm{cm}}^3$.

The radius of the earth is, $r_E=6.37\times {10}^6 \mathrm{m}$.

(a) Determination of the radius of the planet

The volume of the planet is given by,

$V_P=\frac{M_P}{{\rho }_P}$

Here, $M_P$ is the planet's mass and ${\rho }_P$ is the planet's density.

Substitute the values in formula of, $V_P$

$\begin{array}{rcl}{V}_P& =& \frac{3.28\times {10}^{25} \mathrm{kg}}{1.76 \mathrm{g}/{\mathrm{cm}}^3}\left(\frac{1 \mathrm{g}/{\mathrm{cm}}^3}{1000 \mathrm{kg}/{\mathrm{m}}^3}\right)\\ & =& 1.86\times {10}^{22}{ \mathrm{m}}^3\end{array}$

Consider the planet as a sphere, it's volume can be expressed as,

$V_P=\frac{4\pi }{3}{r}_P^3$

Here, $r_P$ is the planet's radius.

Solution of the above equation for ${\Gamma }_{\mathrm{P}}$ gives,

$r_P={\left(\frac{3V_P}{4\pi }\right)}^{1/3}$

Substitute all the values in the above,

$\begin{array}{rcl}{r}_P& =& {\left(\frac{3\times \left(1.86\times {10}^{22}{ \mathrm{m}}^3\right)}{4\pi }\right)}^{1/3}\\ & =& 1.64\times {10}^7 \mathrm{m}\left(\frac{1 \mathrm{km}}{1000 \mathrm{m}}\right)\\ & =& 1.64\times {10}^4 \mathrm{km}\end{array}$

Thus, the planet's radius is $1.64\times {10}^4 \mathrm{km}$.

(b) determination of the radius of the planet as multiple of earth’s radius

The planet’s radius is given by,

$r_P=1.64\times {10}^4 \mathrm{km}$

The radius of the planet in terms of earth's radius can be expressed as,

$r_P=\left(\frac{r_P\times r_E}{r_E}\right)$

Here, $r_E$ is the radius of the earth whose value is $r_E=6.37\times {10}^6 \mathrm{m}$.

Substitute values in the above equation.

$\begin{array}{rcl}{r}_P& =& \left(\frac{1.64\times {10}^7 \mathrm{m}\times r_E}{6.37\times {10}^6 \mathrm{m}}\right)\\ & =& 2.57r_E\end{array}$

Thus, the planet's radius in terms of the earth's radius $\left(r_E\right)$ is $2.57r_E$.

Final Solution

The planet's radius is $1.64\times {10}^4 \mathrm{km}$ and in terms of the earth's radius $\left(r_E\right)$, it is $2.57r_E$.