Gravitational Field of a Point Mass

Gravitational Field Strength at a Point

The gravitational field strength at a point is a physical quantity that describes the intensity of a gravitational field at a specific location. It is defined as the gravitational force per unit mass acting on a small test mass placed at that specific point. Mathematically, this relationship is expressed by the equation $g = \frac{F}{m}$, where $g$ represents the gravitational field strength measured in newtons per kilogram ($N\,kg^{-1}$), $F$ is the gravitational force in newtons ($N$), and $m$ is the mass of the object situated within the gravitational field measured in kilograms ($kg$).

Derivation of Gravitational Field Strength for a Point Mass

The gravitational field strength due to a point mass located within a gravitational field can be derived by combining the general definition of field strength with Newton’s law of gravitation. Newton’s law of gravitation state that the attractive force $F_G$ between two masses, $M$ and $m$, separated by a distance $r$, is given by the equation $F_G = \frac{GMm}{r^2}$. By rearranging the basic definition of gravitational field strength to make force the subject, we obtain $F = mg$. By equating the gravitational force from Newton's law with the force due to field strength ($F = F_G$), we get the expression $mg = \frac{GMm}{r^2}$. Canceling the test mass $m$ from both sides of the equation yields the final formula for the gravitational field strength due to a point mass: $g = \frac{GM}{r^2}$. In this equation, $G$ represents Newton’s Gravitational Constant, $M$ is the mass of the primary body producing the gravitational field, and $r$ is the distance between the point source and the position in the field.

Inverse Square Law and Vector Direction

Gravitational field strength $g$ is a vector quantity, meaning it has both magnitude and direction. The direction of $g$ is always oriented towards the center of the mass producing the field, which corresponds to the direction of the gravitational field lines. There is an inverse square law relationship between the gravitational field strength $g$ and the orbital radius $r$, which is expressed as $g \propto \frac{1}{r^2}$. This implies that as the distance from the center of the mass increases, the field strength decreases significantly; specifically, if the distance doubles, the field strength becomes one-fourth of its original value.

Examiner Tips for Point Mass Situations

It is essential for students to distinguish between the two primary contexts in which gravitational field strength is applied. The first context is the general definition of gravitational field strength at a point due to an object creating the field, defined as $g = \frac{F}{m}$. The second context specifically addresses the gravitational field strength at a distance $r$ from a point mass (or a larger body treated as a point mass), which is calculated using $g = \frac{GM}{r^2}$. In the latter case, it is critical to remember that $g$ decreases as the distance $r$ increases by a factor of $\frac{1}{r^{2}}$.

Relationship Between Density and Gravitational Field Strength

Gravitational field strength can also be expressed in terms of the density and volume of the body producing the field. Given the density equation $\rho = \frac{M}{V}$, the mass of a body can be described as $M = \rho V$. When a planetary body is approximated as a sphere, its volume is given by $V = \frac{4}{3}\pi r^3$. Substituting these into the gravity equation, we get $g = \frac{G\rho V}{r^2}$. Further substituting the volume of a sphere results in $g = \frac{G\rho(\frac{4\pi r^3}{3})}{r^2}$, which simplifies to the expression $g = \frac{4\pi G\rho r}{3}$. This formula shows that for a body of constant density, the surface gravity is directly proportional to its radius and density.

Worked Example: Moon and Earth Radii Ratio

Consider a scenario where the mean density of the Moon is $\frac{3}{5}$ the mean density of the Earth, and the gravitational field strength at the Moon's surface is $\frac{1}{6}$ that of the Earth's surface. To determine the ratio of the Moon’s radius $r_M$ to the Earth’s radius $r_E$, we use the integrated formula for field strength. First, identify the known ratios: $\rho_M = \frac{3}{5}\rho_E$ and $g_M = \frac{1}{6}g_E$. Using the relationship $g = \frac{4\pi G\rho r}{3}$, the ratio of the field strengths is $\frac{g_M}{g_E} = \frac{\rho_M r_M}{\rho_E r_E}$. Rearranging for the radius ratio gives $\frac{r_M}{r_E} = \frac{\rho_E g_M}{\rho_M g_E}$. Substituting the given values, we find $\frac{r_M}{r_E} = \frac{\rho_E (\frac{1}{6}g_E)}{(\frac{3}{5}\rho_E) g_E} = \frac{1}{6} \div \frac{3}{5} = \frac{5}{18}$. This results in a ratio of approximately $0.28$.

The Value of g on the Earth's Surface

On Earth, the gravitational field strength $g$ is approximately constant for relatively small changes in height near the surface. Within the Earth's atmosphere, this value is taken as $g = 9.81\,N\,kg^{-1}$. The approximation of $g$ as a constant is valid because the radius of the Earth, denoted as $R$, is significantly larger than the distance between the surface and an object in the atmosphere, denoted as $h$. Mathematically, $R \gg h$. For instance, for an object orbiting at a height of $150\,km$ ($1.5 \times 10^5\,m$), and given the Earth's radius is $6400\,km$ ($6.4 \times 10^6\,m$), the total orbital radius $r$ is $6.55 \times 10^6\,m$. Because $6.4 \times 10^6 \cong 6.55 \times 10^6$, the orbital radius is roughly equal to the Earth's radius, and small changes in height do not significantly affect the total height $h$. Thus, the field strength at height $h$ is $g = \frac{GM}{(R + h)^2} \simeq \frac{GM}{R^2}$.

Worked Example: Gravity at the Peak of Mount Everest

To demonstrate the small effect of altitude on gravity, consider Mount Everest, which has a peak at $8850\,m$ above Earth's surface. Using the Earth's mass ($M = 5.97 \times 10^{24}\,kg$) and radius ($R = 6370\,km$), we can calculate the percentage decrease in $g$. First, calculate $g$ at the surface ($g_0$): $g_0 = \frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24})}{(6370 \times 10^3)^2} = 9.8134\,N\,kg^{-1}$. Next, determine the value of $r$ at the peak: $r = 6370 + 8.85 = 6378.85\,km = 6378.85 \times 10^3\,m$. The field strength at this height ($g'$) is $g' = \frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24})}{(6378.85 \times 10^3)^2} = 9.7862\,N\,kg^{-1}$. The percentage decrease is calculated as $\frac{g_0 - g'}{g_0} \times 100\%$. Substituting the values, $\frac{9.8134 - 9.7862}{9.8134} \times 100 = 0.28\%$. This confirms that gravity at the top of Everest is approximately $0.3\%$ less than at the Earth's surface.