Question

For a spherical planet with mass $\mathbf{M}$, volume $\mathbf{V}$, and radius $\mathbf{R}$,derive an expression for the acceleration due to gravity at the planet’s surface, $\mathbf{g}$, in terms of the average density of the planet, $\mathrm{\rho }=\mathrm{M}/\mathrm{V}$, and the planet’s diameter, $\mathbf{D}\mathbf{=}\mathbf{2}\mathbf{R}$. The table gives the values of $\mathbf{D}$and $\mathbf{g}$for the eight major planets:

(a) Treat the planets as spheres. Your equation for as a function of and shows that if the average density of the planets is constant, a graph of versus will be well represented by a straight line. Graph as a function of for the eight major planets. What does the graph tell you about the variation in average density? (b) Calculate the average density for each major planet. List the planets in order of decreasing density, and give the calculated average density of each. (c) The earth is not a uniform sphere and has greater density near its center. It is reasonable to assume this might be true for the other planets. Discuss the effect this nonuniformity has on your analysis. (d) If Saturn had the same average density as the earth, what would be the value of at Saturn’s surface?

Short answer

a) The graph of $\mathrm{g}$versus$\mathrm{D}$ will change with the change in the average density.

b)The average density of the planets in the decreasing order are of the planets Mercury, Mars, Venus, Earth, Neptune, Uranus, Jupiter and Saturn which are $5447.85\mathrm{kg}/{\mathrm{m}}^3$, $3911.89\mathrm{kg}/{\mathrm{m}}^3$, $2192.59\mathrm{kg}/{\mathrm{m}}^3$, $2079.62\mathrm{kg}/{\mathrm{m}}^3$, $1591.32\mathrm{kg}/{\mathrm{m}}^3$, $1219.62\mathrm{kg}/{\mathrm{m}}^3$, $1157.65\mathrm{kg}/{\mathrm{m}}^3$,$535.05\mathrm{kg}/{\mathrm{m}}^3$ respectively.

c)The eight major planets have also variable average densities due to their spin about their axis.

d) The value of the$\mathrm{g}$ at the Saturn’s surface is$34.99\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Identification of given data

Given data:

• The spherical planet has mass $\text{M}$.
• The spherical planet has volume $\mathrm{V}$.
• The spherical planet has radius $\mathrm{R}$.
• The spherical planet has diameter $\mathrm{D}=2\mathrm{R}$.

Significance of the gravity equation

The acceleration due to gravity of a planet is the ratio of the product of the Universal Gravitational constant and mass of the planet to the square of their radius.

Determination of the acceleration due to gravity

The equation of the density of the planet is expressed as-

$$\begin{gathered} \mathrm{\rho }=\frac{\mathrm{M}}{\mathrm{V}} \\ \mathrm{M}=\mathrm{\rho V} \end{gathered}$$

Here,$\mathrm{\rho }$is the density of the planet,$\mathrm{M}$ is the mass of the planet and$\mathrm{V}$ is the volume of the planet.

The equation of the acceleration due to gravity is expressed as-

$\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}$

Here,$\mathrm{g}$ is the acceleration due to gravity, $\mathrm{G}$is the gravitational constant and $\mathrm{R}$is the radius of the spherical planet.

For$\mathrm{M}=\mathrm{\rho V}$ and $\mathrm{D}=2\mathrm{R}$,

$\begin{array}{rcl}\mathrm{g}& =& \frac{\mathrm{G\rho V}}{{\left({\displaystyle \frac{\mathrm{D}}{2}}\right)}^2}\\ & =& \frac{4\mathrm{G\rho V}}{{\mathrm{D}}^2}...........................................\left(1\right)\\ & & \end{array}$

Determination of the variation in average densitya)

From the expression of the acceleration due to gravity, the graph of $\mathrm{g}$as a function $\mathrm{D}$of has been described below-

This graph is applicable for all the eight major planets if the average density of the planets is constant. However, if the average density of the planet changes, then the graph will also change.

Thus, the graph of$\mathrm{g}$ versus$\mathrm{D}$ will change with the change in the average density.

Determination of the average density of the eight major planetsb)

From the equation $\left(1\right)$the equation of the average density of the planets is expressed as-

$\mathrm{\rho }=\frac{{\mathrm{gD}}^2}{4\mathrm{GV}}............................\left(2\right)$

As the planets are spherical in nature, then the equation of the volume of the planet can be expressed as-

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

Then equation $\left(2\right)$becomes-

$\begin{array}{rcl}\mathrm{\rho }& =& \frac{{\mathrm{gD}}^2}{4\mathrm{G}\times \left({\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3\right)}\\ & =& \frac{3{\mathrm{gD}}^2}{16{\mathrm{G\pi R}}^3}\end{array}$

For $\mathrm{D}=2\mathrm{R}$,

$\begin{array}{rcl}\rho & =& \frac{3g{D}^2}{16G\mathrm{\pi }{\left({\displaystyle \frac{D}{2}}\right)}^3}\\ & =& \frac{24g}{16G\mathrm{\pi }D}........................................\left(3\right)\end{array}$

For the planet Mercury,

Substituting $\mathrm{D}=4879\mathrm{km}$,$\mathrm{G}=6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2$and$\mathrm{g}=3.7\mathrm{m}/{\mathrm{s}}^2$in the equation 3),

$\begin{array}{rcl}\mathrm{\rho }& =& \frac{24\times 3.7\mathrm{m}/{\mathrm{s}}^2}{16\times 6.7\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 3.14\times 4879\mathrm{km}}\\ & =& \frac{24\times 3.7\mathrm{m}/{\mathrm{s}}^2}{3.35\times {10}^{-09}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\times 4879\times {10}^3\mathrm{m}}\\ & =& \frac{24\times 3.7\mathrm{m}/{\mathrm{s}}^2}{0.0163{\mathrm{Nm}}^3/{\mathrm{kg}}^2}\\ & =& 5447.85{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\end{array}$

Hence, further as,

$\begin{array}{rcl}\mathrm{\rho }& =& 5447.85{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\\ & =& 5447.85{\mathrm{kg}}^2.\mathrm{m}/{\mathrm{m}}^3.{\mathrm{s}}^2\times \frac{1}{\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\\ & =& 5447.85\mathrm{kg}/{\mathrm{m}}^3\end{array}$

For the planet Venus,

Substituting$\mathrm{D}=12104\mathrm{km}$, $\mathrm{G}=6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2$and$\mathrm{g}=8.9\mathrm{m}/{\mathrm{s}}^2$in the equation $\left(3\right)$,

$\begin{array}{rcl}\mathrm{\rho }& =& \frac{24\times 8.9\mathrm{m}/{\mathrm{s}}^2}{16\times 6.017\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 3.14\times 12014\mathrm{km}}\\ & =& \frac{24\times 8.9\mathrm{m}/{\mathrm{s}}^2}{3.35\times {10}^{-09}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 12014\times {10}^3\mathrm{m}}\\ & =& \frac{24\times 3.7\mathrm{m}/{\mathrm{s}}^2}{0.0405\mathrm{N}.{\mathrm{m}}^3/{\mathrm{kg}}^2}\\ & =& 2192.59{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\end{array}$

Hence, further as,

$$\begin{gathered} \mathrm{\rho }=2192.59{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2 \\ =2192.59{\mathrm{kg}}^2.\mathrm{m}/{\mathrm{m}}^3.{\mathrm{s}}^2\times \frac{1}{\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =2192.59\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

For the planet Earth,

Substituting $\mathrm{D}=12756\mathrm{km}$,$\mathrm{G}=6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2$and$\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$in the equation

$\left(3\right)$

localid="1655123018356" $\begin{array}{rcl}\mathrm{\rho }& =& \frac{24\times 9.8\mathrm{m}/{\mathrm{s}}^2}{16\times 6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 3.14\times 6792\mathrm{km}}\\ & =& \frac{24\times 9.8\mathrm{m}/{\mathrm{s}}^2}{3.35\times {10}^{-09}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 6792\times {10}^3\mathrm{m}}\\ & =& \frac{24\times 9.8\mathrm{m}/{\mathrm{s}}^2}{0.0427\mathrm{N}.{\mathrm{m}}^3/{\mathrm{kg}}^2}\\ & =& 2079.62.{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\end{array}$

Hence, further as,

localid="1655122399974" $\begin{array}{rcl}\mathrm{\rho }& =& 2079.62{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\\ & =& 2079.62{\mathrm{kg}}^2.\mathrm{m}/{\mathrm{m}}^3.{\mathrm{s}}^2\times \frac{1}{\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\\ & =& 2079.62\mathrm{kg}/{\mathrm{m}}^3\end{array}$

For the planet Mars,

Substituting$D=6792km$,$\mathrm{G}=6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2$and$\mathrm{g}=3.7\mathrm{m}/{\mathrm{s}}^2$in the equation$\left(3\right)$

$\begin{array}{rcl}\mathrm{\rho }& =& \frac{24\times 3.7\mathrm{m}/{\mathrm{s}}^2}{16\times 6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 3.14\times 6792\mathrm{km}}\\ & =& \frac{24\times 3.7\mathrm{m}/{\mathrm{s}}^2}{3.35\times {10}^{-09}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 6792\times {10}^3\mathrm{m}}\\ & =& \frac{24\times 3.7\mathrm{m}/{\mathrm{s}}^2}{0.0227\mathrm{N}.{\mathrm{m}}^3/{\mathrm{kg}}^2}\\ & =& 3911.89k{g}^2.m/N.m^3.s^2\end{array}$

Hence, further as,

$\begin{array}{rcl}\mathrm{\rho }& =& 3911.89{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\\ & =& 3911.89{\mathrm{kg}}^2.\mathrm{m}/{\mathrm{m}}^3.{\mathrm{s}}^2\times \frac{1}{\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\\ & =& 3911.89\mathrm{kg}/{\mathrm{m}}^3\end{array}$

For the planet Jupiter,

Substituting$\mathrm{D}=142,984\mathrm{km}$,$\mathrm{G}=6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2$and$\mathrm{g}=23.1\mathrm{m}/{\mathrm{s}}^2$in the equation (3),

$\begin{array}{rcl}\mathrm{\rho }& =& \frac{24\times 23.1\mathrm{m}/{\mathrm{s}}^2}{16\times 6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 3.14\times 142,984\mathrm{km}}\\ & =& \frac{24\times 23.1\mathrm{m}/{\mathrm{s}}^2}{3.35\times {10}^{-09}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 142,984\times {10}^3\mathrm{m}}\\ & =& \frac{24\times 23.1\mathrm{m}/{\mathrm{s}}^2}{0.4789\mathrm{N}.{\mathrm{m}}^3/{\mathrm{kg}}^2}\\ & =& 1157.65{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{kg}}^2\end{array}$

Hence, further as,

$\begin{array}{rcl}\mathrm{\rho }& =& 1157.65{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\\ & =& 1157.65{\mathrm{kg}}^2.\mathrm{m}/{\mathrm{m}}^3.{\mathrm{s}}^2\times \frac{1}{\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\\ & =& 1157.65\mathrm{kg}/{\mathrm{m}}^3\end{array}$

For the planet Saturn,

Substituting$\mathrm{D}=120536\mathrm{km}$,$\mathrm{G}=6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2$and$\mathrm{g}=9.0\mathrm{m}/{\mathrm{s}}^2$in the equation (3),

$\begin{array}{rcl}\mathrm{\rho }& =& \frac{24\times 9.0\mathrm{m}/{\mathrm{s}}^2}{16\times 6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 3.14\times 120536\mathrm{km}}\\ & =& \frac{24\times 9.0\mathrm{m}/{\mathrm{s}}^2}{3.35\times {10}^{-09}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 120536\times {10}^3\mathrm{m}}\\ & =& \frac{24\times 9.0\mathrm{m}/{\mathrm{s}}^2}{0.4037\mathrm{N}.{\mathrm{m}}^3/{\mathrm{kg}}^2}\\ & =& 535.05{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\end{array}$

Hence, further as,

$\begin{array}{rcl}\mathrm{\rho }& =& 535.05{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\\ & =& 535.05{\mathrm{kg}}^2.\mathrm{m}/{\mathrm{m}}^3.{\mathrm{s}}^2\times \frac{1}{\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\\ & =& 535.05\mathrm{kg}/{\mathrm{m}}^3\end{array}$

For the planet Uranus,

Substituting$\mathrm{D}=51118\mathrm{km}$,$G=6.67\times {10}^{-11}N.m^2/k{g}^2$and $\mathrm{g}=8.7\mathrm{m}/{\mathrm{s}}^2$in the equation (3),

$\begin{array}{rcl}\mathrm{\rho }& =& \frac{24\times 8.7\mathrm{m}/{\mathrm{s}}^2}{16\times 6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 3.14\times 51118\mathrm{km}}\\ & =& \frac{24\times 8.7\mathrm{m}/{\mathrm{s}}^2}{3.35\times {10}^{-09}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 51118\times {10}^3\mathrm{km}}\\ & =& \frac{24\times 8.7\mathrm{m}/{\mathrm{s}}^2}{0.1712\mathrm{N}.{\mathrm{m}}^3/{\mathrm{kg}}^2}\\ & =& 1219.62{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\end{array}$

Hence, further as,

$\begin{array}{rcl}\mathrm{\rho }& =& 1219.62{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\\ & =& 1219.62{\mathrm{kg}}^2.\mathrm{m}/{\mathrm{m}}^3.{\mathrm{s}}^2\times \frac{1}{\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\\ & =& 1219.62\mathrm{kg}/{\mathrm{m}}^3\end{array}$

For the planet Neptune,

Substituting $\mathrm{D}=49528\mathrm{km}$, $\mathrm{G}=6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2$and $\mathrm{g}=11.0\mathrm{m}/{\mathrm{s}}^2$in the equation (3),

$\begin{array}{rcl}\mathrm{\rho }& =& \frac{24\times 11.0\mathrm{m}/{\mathrm{s}}^2}{16\times 6.67\times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 3.14\times 49525\mathrm{km}}\\ & =& \frac{24\times 11.0\mathrm{m}/{\mathrm{s}}^2}{3.35\times {10}^{-09}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\times 49525\times {10}^3\mathrm{m}}\\ & =& \frac{24\times 11.0\mathrm{m}/{\mathrm{s}}^2}{0.1659\mathrm{N}.{\mathrm{m}}^3/{\mathrm{kg}}^2}\\ & =& 1591.32{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\end{array}$

Hence, further as,

$\begin{array}{rcl}\mathrm{\rho }& =& 1591.32{\mathrm{kg}}^2.\mathrm{m}/\mathrm{N}.{\mathrm{m}}^3.{\mathrm{s}}^2\\ & =& 1591.32{\mathrm{kg}}^2.\mathrm{m}/{\mathrm{m}}^3.{\mathrm{s}}^2\times \frac{1}{\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\\ & =& 1591.32\mathrm{kg}/{\mathrm{m}}^3\end{array}$

Thus, the average density of the planets in the decreasing order are of the planets Mercury, Mars, Venus, Earth, Neptune, Uranus, Jupiter and Saturn which are $5447.85\mathrm{kg}/{\mathrm{m}}^3$, $3911.89\mathrm{kg}/{\mathrm{m}}^3$, $2192.59\mathrm{kg}/{\mathrm{m}}^3$, $2079.62\mathrm{kg}/{\mathrm{m}}^3$, $1591.32\mathrm{kg}/{\mathrm{m}}^3$, $1219.62\mathrm{kg}/{\mathrm{m}}^3$, $1157.65\mathrm{kg}/{\mathrm{m}}^3$, $535.05\mathrm{kg}/{\mathrm{m}}^3$respectively.

Determination of the effect of the nonuniformityc)

Along with the planet Earth, the other eight major planets are also not uniform as they spin around their axis and around the sun. However, the force of spin mainly acts against the gravity and it eventually causes the planets to more bulge out around their respective equator.

Thus, due to the spinning force, the eight major planets have also variable average densities.

Determination of the value of acceleration due to gravity at Saturn’s surfaced)

If the average density of Saturn is same as of the Earth, then from the equation 3), the equation of the acceleration due to gravity of Saturn can be expressed as,

For,and,

Hence, further as,

Thus, the value of the at the Saturn’s surface is.