Solved Problem on Kepler's Laws and Gravitation

Variations in the gravitational field at the Earth's surface can arise from irregularities in the distribution of its mass. Consider the Earth as a sphere of uniform radius R and density ρ, with a spherical cavity of radius a, entirely contained within it. The distance between the centers O, of the Earth, and C, of the cavity, is d, which can vary from 0 (zero) to R−a, thus causing a variation in the gravitational field at a point P, on the surface of the Earth, aligned with O and C (see figure). If G₁ is the gravitational force at P without the existence of the cavity on Earth, and G₂, is the gravitational force at the same point, considering the existence of the cavity. What will be the maximum value of the relative variation: \( \left(G_{1}-G_{2}\right)/G_{1} \), obtained by moving the cavity position?

Problem data:

Radius of the Earth: R;
Density of the Earth: ρ;
Radius of the inner cavity contained in the Earth: a;
Distance between the center O of the Earth and the center C of the cavity: d.

Problem diagram:

The figure shows the Earth in a section with a spherical cavity inside.

Figure 1

Solution

The gravitational field of the Earth without a cavity

The magnitude of the gravitational field, G₁, of the Earth without the cavity at a point P situated at a distance R from the center is given by (Figure 2)

\[ \begin{gather} G_{1}=\frac{GM}{R^{2}} \tag{I} \end{gather} \]

where G is the universal gravitational constant, M is the mass of the Earth, and V is the volume of the Earth, mass is given by

\[ \begin{gather} M=\rho V \tag{II} \end{gather} \]

Figure 2

the problem considers the Earth as a sphere, the volume of a sphere is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {V=\frac{4}{3}\pi r^{3}} \tag{III} \end{gather} \]

for r = R

\[ \begin{gather} V=\frac{4}{3}\pi R^{3} \tag{IV} \end{gather} \]

substituting expression (IV) into expression (II), the mass is given by

\[ \begin{gather} M=\rho \frac{4}{3}\pi R^{3} \tag{V} \end{gather} \]

substituting expression (V) into expression (I), the strength of the Earth's gravitational field will be

\[ G_{1}=\frac{G\rho \frac{4}{3}\pi R^{\cancel{3}}}{\cancel{R^{2}}} \]

canceling the values of R³ in the numerator and R² in the denominator

\[ \begin{gather} G_{1}=\frac{4}{3}\pi RG\rho \tag{VI} \end{gather} \]

The gravitational field produced by the sphere removed from the cavity

The gravitational field, G_S, produced by a sphere of radius a, mass m, volume v, and the same density ρ as the Earth, at a point P at a distance (R−d) will be (Figure 3)

\[ \begin{gather} G_{S}=\frac{Gm}{(R-d)^{2}} \tag{VII} \end{gather} \]

the mass of the sphere is calculated by

Figure 3

\[ \begin{gather} m=\rho v \tag{VIII} \end{gather} \]

setting r = a in expression (III)

\[ \begin{gather} v=\frac{4}{3}\pi a^{3} \tag{IX} \end{gather} \]

substituting expression (IX) into expression (VIII), the mass will be

\[ \begin{gather} m=\rho \frac{4}{3}\pi a^{3} \tag{X} \end{gather} \]

substituting expression (X) into expression (VII), the field of the sphere will be

\[ \begin{gather} G_{S}=\frac{\dfrac{4}{3}\pi a^{3}G\rho }{(R-d)^{2}}\\ G_{S}=\frac{4}{3}\pi a^{3}G\rho\frac{1}{(R-d)^{2}} \tag{XI} \end{gather} \]

Earth's gravitational field with cavity

The gravitational field G₂ produced by the Earth, at a point P, with the cavity left when removing a sphere from its interior will be given by the total field given in G₁ (equation V) minus the field G_S of the sphere removed from its interior (equation X)

\[ G_{2}=G_{1}-G_{S}=\frac{4}{3}\pi RG\rho -\frac{4}{3}\pi a^{3}G\rho\frac{1}{(R-d)^{2}} \]

factoring the term \( \frac{4}{3}\pi G\rho \) on the right-hand side of the equation

\[ \begin{gather} G_{2}=\frac{4}{3}\pi G\rho \left[R-\frac{a^{3}}{(R-d)^{2}}\right] \tag{XII} \end{gather} \]

That is the expression of the Earth's gravitational field with a cavity inside it at point P (Figure 4).
Using expressions (VI) and (XII), the required variation in the problem is calculated

\[ \frac{G_{1}-G_{2}}{G_{1}}=\frac{\cancel{\dfrac{4}{3}}\cancel{\pi} R \cancel{G}\cancel{\rho} -\cancel{\dfrac{4}{3}}\cancel{\pi} \cancel{G}\cancel{\rho} \left[R-\dfrac{a^{3}}{(R-d)^{2}}\right]}{\cancel{\dfrac{4}{3}}\cancel{\pi} R \cancel{G}\cancel{\rho}} \]

canceling the term \( \frac{4}{3}\pi G\rho \) which appears in all terms of the expression

\[ \begin{gather} \frac{G_{1}-G_{2}}{G_{1}}=\frac{R-R+\dfrac{a^{3}}{(R-d)^{2}}}{R}\\[5pt] \frac{G_{1}-G_{2}}{G_{1}}=\frac{a^{3}}{R(R-d)^{2}} \end{gather} \]

Figure 4

The expression above gives the relative variation of the gravitational field and will have a maximum value when the denominator has a minimum value. Since R (the radius of the Earth is a constant) then d must be maximum to make the difference (R−d) minimum, the problem tells us that d varies from zero (minimum value) to R−a (maximum value), so d = R−a

\[ \begin{gather} \frac{G_{1}-G_{2}}{G_{1}}=\frac{a^{3}}{R[R-(R-a)]^{2}}\\[5pt] \frac{G_{1}-G_{2}}{G_{1}}=\frac{a^{3}}{R[R-R+a]^{2}}\\[5pt] \frac{G_{1}-G_{2}}{G_{1}}=\frac{a^{\cancel{3}}}{R\cancel{[a]^{2}}} \end{gather} \]

canceling the terms a³ in the numerator, and a² in the denominator, the maximum variation of the field is given by

\[ \bbox[#FFCCCC,10px] {\frac{G_{1}-G_{2}}{G_{1}}=\frac{a}{R}} \]

Note: The expression (I) is obtained from Newton's Second Law

\[ \bbox[#99CCFF,10px] {F=m a} \]

and the Law of Universal Gravitation

\[ \bbox[#99CCFF,10px] {F=G\frac{Mm}{r^{2}}} \]

Setting a = g in the first expression and equating the two expressions

\[ \begin{gather} m g=G\frac{Mm}{r^{2}}\\ g=\frac{GM}{r^{2}} \end{gather} \]

this expression gives the acceleration due to gravity for all bodies, given their mass M and radius r.

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .