Question

the planet jupiter has a mass of $1.9\times10^{27}$ kg and a radius of 72,000 km. the earth, meanwhile, has a mass of $6.0\times10^{24}$ kg and a radius of 6,400 km.
part (a)
what is the volume $v$ of jupiter in $m^{3}$?
$v = 1.56\times10^{24}$
$v = 1.560\times10^{24}m^{3}$
part (b)
calculate the density $
ho$ of jupiter in $kg/m^{3}$.
$
ho=square kg/m^{3}$

Explanation:

Step1: Recall the density formula

Density formula is \(
ho=\frac{m}{V}\), where \(m\) is mass and \(V\) is volume.

Step2: Substitute the values

Given \(m = 1.9\times10^{27}\text{ kg}\) and \(V=1.560\times 10^{24}\text{ m}^3\). Then \(
ho=\frac{1.9\times 10^{27}}{1.560\times 10^{24}}\).

Step3: Calculate the value

Using the rule of exponents \(\frac{a^m}{a^n}=a^{m - n}\), we have \(\frac{1.9}{1.560}\times10^{27-24}\). \(\frac{1.9}{1.560}\approx1.22\), so \(
ho\approx1.22\times 10^{3}\text{ kg/m}^3\).

Answer:

\(1.22\times 10^{3}\text{ kg/m}^3\)