Comprehensive Study Guide on Uniform Circular Motion and Universal Gravitation

Kinematics of Uniform Circular Motion

Uniform circular motion occurs when an object travels along a circular path at a constant speed. While the scalar speed remains unchanging, the direction of the velocity vector continuously rotates, meaning the object constantly undergoes acceleration directed toward the center of rotation.

The time needed to execute one complete revolution is the period, denoted by $T$. The number of complete cycles per unit of time is defined as the rotational frequency $f$, which is the reciprocal of the period:

$f = \frac{1}{T}$

The orbital speed (or tangential velocity) $v$ represents the arc length traveled during one period divided by the period itself:

$v = \frac{2\pi \cdot r}{T} = 2\pi \cdot r \cdot f$

The rate at which the angular position changes over time is known as the angular velocity or angular frequency, denoted by $\omega$:

$\omega = \frac{\Delta \varphi}{\Delta t} = \frac{2\pi}{T} = 2\pi \cdot f$

The SI unit of angular velocity is expressed as:

$[\omega] = \frac{\text{rad}}{\text{s}}$

By substituting $\omega$ into the equation for tangential speed, the direct coupling between linear velocity and angular velocity is established:

$v = \omega \cdot r$

Because the direction of the velocity vector changes continuously toward the rotation axis, an object in uniform circular motion experiences centripetal acceleration, denoted by $a_z$. Its magnitude can be expressed through angular velocity, tangential velocity, and trajectory radius:

$a_z = \omega \cdot v = \omega^2 \cdot r = \frac{v^2}{r}$

Applications and Quantitative Analysis in Circular Kinematics

The kinematics of circular motion govern phenomena ranging from planetary orbits to laboratory separation techniques and atmospheric events.

The Earth travels practically uniformly on an approximately circular orbit around the Sun at an average orbital distance of $r = 1.496 \times 10^{11}\,\text{m}$. The distance $s$ traversed by the Earth during one full revolution around the Sun is the circumference of this orbit:

$s = 2\pi \cdot r = 2\pi \cdot (1.496 \times 10^{11}\,\text{m}) = 9.400 \times 10^{11}\,\text{m}$

Given an orbital period of one year, defined as $T = 365.25\,\text{d} = 3.156 \times 10^7\,\text{s}$, the orbital speed of the Earth is calculated as:

$v = \frac{2\pi \cdot r}{T} = \frac{2\pi \cdot (1.496 \times 10^{11}\,\text{m})}{3.156 \times 10^7\,\text{s}} = 29.79\,\text{km/s}$

The resulting centripetal acceleration holding the Earth in its path is:

$a = \frac{v^2}{r} = \frac{(29.79 \times 10^3\,\text{m/s})^2}{1.496 \times 10^{11}\,\text{m}} = 5.930 \times 10^{-3}\,\text{m/s}^2$

In human aerospace physiology, large centrifuges accommodating a person are deployed to simulate the intense accelerations experienced during rocket launches. To generate a simulated acceleration of $a_z = 3\,g$ (where standard gravitational acceleration is taken as $g = 9.81\,\text{m/s}^2$) at an operational arm radius of $r = 10\,\text{m}$, the required frequency is derived from the acceleration relation:

$a_z = \frac{v^2}{r} = 4\pi^2 \cdot r \cdot f^2$

$f = \sqrt{\frac{a_z}{4\pi^2 \cdot r}} = \sqrt{\frac{3 \cdot 9.81\,\text{m/s}^2}{4\pi^2 \cdot 10\,\text{m}}} = 0.27\,\text{Hz}$

Converting this frequency to revolutions per minute yields:

$0.27\,\text{s}^{-1} \times 60\,\text{s/min} = 16\,\text{min}^{-1}$

In 1925, Theodor Svedberg invented the analytical ultracentrifuge, an achievement recognized by the Nobel Prize in Chemistry in 1926. Modern ultracentrifuges spin rapidly enough to subject biological samples to centripetal accelerations reaching up to $a = 10^6\,g$. For a test sample positioned at a radius of $r = 5.0\,\text{cm} = 0.05\,\text{m}$, the operational rotational frequency must be:

$f = \sqrt{\frac{a}{4\pi^2 \cdot r}} = \sqrt{\frac{10^6 \cdot 9.81\,\text{m/s}^2}{4\pi^2 \cdot 0.05\,\text{m}}} = 2.2\,\text{kHz}$

Because centripetal acceleration depends on the parameters according to $a = 4\pi^2 \cdot r \cdot f^2$, doubling the rotation frequency causes a fourfold ($4\times$) increase in acceleration due to the quadratic dependency ($a \propto f^2$), whereas doubling the radius results in a twofold ($2\times$) increase due to linear scaling ($a \propto r$).

Large-scale atmospheric phenomena also exhibit rotational dynamics. Consider a tropical cyclone modeled as a rigid, rotating air mass possessing a uniform angular velocity $\omega$ across its entire structure. A meteorological station positioned $r_1 = 500\,\text{km} = 500 \times 10^3\,\text{m}$ from the eye records a peripheral wind velocity of $v_1 = 230\,\text{km/h} = 63.9\,\text{m/s}$. The centripetal acceleration experienced by the air at this outer boundary is:

$a = \frac{v_1^2}{r_1} = \frac{(63.9\,\text{m/s})^2}{500 \times 10^3\,\text{m}} = 8.16 \times 10^{-3}\,\text{m/s}^2$

The rotational period of the entire cyclone mass is:

$T = \frac{2\pi \cdot r_1}{v_1} = \frac{2\pi \cdot (500 \times 10^3\,\text{m})}{63.9\,\text{m/s}} = 4.92 \times 10^4\,\text{s} = 13.7\,\text{h}$

Because angular velocity is constant throughout the rigid body ($\omega = \text{const}$), tangential speed scales directly with radial distance. At an intermediate distance of $r_2 = 250\,\text{km} = 250 \times 10^3\,\text{m}$, the local wind speed is:

$v_2 = \frac{2\pi \cdot r_2}{T} = \frac{r_2 \cdot v_1}{r_1} = \frac{(250 \times 10^3\,\text{m}) \cdot (63.9\,\text{m/s})}{500 \times 10^3\,\text{m}} = 31.9\,\text{m/s} = 115\,\text{km/h}$

In vehicular safety, a critical problem occurs when a driver unexpectedly encounters an obstacle ahead and must choose between emergency straight-line braking and initiating an evasive $180^\circ$ turn. Assuming the tire-road interface provides the same maximum deceleration in braking as it does centripetal acceleration in cornering ($a = a_z$):

The stopping distance under linear braking from initial speed $v_0$ to complete rest ($v = 0$) is determined from kinematics:

$v^2 = 2 \cdot a \cdot s + v_0^2 = 0 \implies s = \frac{v_0^2}{2 \cdot a}$

Conversely, executing an evasive circular maneuver requires a turning radius of:

$a_z = \frac{v_0^2}{r} \implies r = \frac{v_0^2}{a_z}$

Comparing the two expressions reveals that $s = \frac{r}{2}$. Under identical acceleration limits, the straight-line braking distance is exactly half the radius of the turn. Consequently, emergency braking affords a significantly greater margin of safety for avoiding an impact than attempting to steer around the hazard.

Centripetal Force, Inertial Frames, and Dynamics

Newton's second law dictates that any acceleration requires an unbalanced net force acting in the direction of the acceleration vector. In uniform circular motion, this force is the centripetal force $F_z$:

$F_z = m \cdot a_z = m \cdot \omega \cdot v = m \cdot \omega^2 \cdot r = m \cdot \frac{v^2}{r}$

The centripetal force is directed toward the instantaneous center of the circular trajectory, perpendicular to the instantaneous velocity vector. It is the net physical inward force responsible for continuously deflecting the object's path into a curve.

An observer situated within the rotating system occupies a non-inertial (accelerated) reference frame. Within this rotating frame, the observer experiences the sensation of an outward inertial force termed the centrifugal force, denoted $\vec{F}'_{ZF}$. Its apparent magnitude equals that of the centripetal force:

$F_{ZF} = m \cdot \omega^2 \cdot r$

The centrifugal force is not a real physical interaction but a fictitious or apparent force arising solely from the inertia of the body resisting rotational acceleration within the non-inertial frame of reference.

When a passenger car drives through a horizontal curve, the real physical centripetal force is supplied entirely by the static friction between the tires and the asphalt. If the car drives at excessive speed, the required centripetal force surpasses the maximum available static frictional force. As a consequence, the circular trajectory can no longer be sustained; the radius of curvature expands uncontrollably and the car slides outward off the roadway.

Consider a driver of mass $m = 70\,\text{kg}$ negotiating a horizontal left-hand curve at $v = 120\,\text{km/h} = 33.3\,\text{m/s}$. If the side bolster of the vehicle's seat exerts an inward normal force of $F_z = 600\,\text{N}$ on the driver toward the center of curvature, the radius of the curve is:

$r = \frac{m \cdot v^2}{F_z} = \frac{70\,\text{kg} \cdot (33.3\,\text{m/s})^2}{600\,\text{N}} = 130\,\text{m}$

In accident reconstruction, a driver named Claudio skidded out of a bend of radius $r = 50\,\text{m}$ and collided with a guardrail. The road conditions were dry, corresponding to a static friction coefficient of $\mu_{HR} = 0.75$. While Claudio claimed he was not speeding, dynamic equilibrium dictates the threshold cornering speed before loss of control:

$F_z = F_{HR} \implies \frac{m \cdot v^2}{r} = \mu_{HR} \cdot m \cdot g$

$v = \sqrt{\mu_{HR} \cdot r \cdot g} = \sqrt{0.75 \cdot 50\,\text{m} \cdot 9.81\,\text{m/s}^2} = 19.2\,\text{m/s} = 69\,\text{km/h}$

Claudio must have been traveling at a minimum speed of $69\,\text{km/h}$ at the moment static friction was overcome.

In vertical circular motion, gravity contributes dynamically to the net force along the radial axis. Consider swinging a bucket filled with water ($m = 4.5\,\text{kg}$) on a $40\,\text{cm}$ rope, where the center of gravity of the bucket describes a vertical circle of radius $r = 1.2\,\text{m}$ at a constant speed of $v = 3.5\,\text{m/s}$.

The constant centripetal force required to sustain this motion is:

$F_z = \frac{m \cdot v^2}{r} = \frac{4.5\,\text{kg} \cdot (3.5\,\text{m/s})^2}{1.2\,\text{m}} = 46\,\text{N}$

The downward gravitational force acting on the bucket is:

$F_g = m \cdot g = 4.5\,\text{kg} \cdot 9.81\,\text{m/s}^2 = 44.1\,\text{N}$

At the zenith (top of the loop), gravity points toward the center and helps provide the centripetal acceleration. The tension force $F_{s,\text{top}}$ exerted by the rope is therefore:

$F_{s,\text{top}} = F_z - F_g = 46\,\text{N} - 44.1\,\text{N} = 1.9\,\text{N}$

At the nadir (bottom of the loop), the rope tension must support the entire weight of the bucket in addition to supplying the inward centripetal acceleration:

$F_{s,\text{bottom}} = F_z + F_g = 46\,\text{N} + 44.1\,\text{N} = 100.1\,\text{N}$

Experimental Determination of Smartphone Accelerometer Position

Modern smartphones contain integrated micro-electro-mechanical systems (MEMS) accelerometers. In gaming and sensor applications (such as detecting device orientation and movement in Pokémon Go), knowing the exact coordinate position of the accelerometer chip inside the phone casing is vital. Because manufacturers do not publish these internal hardware coordinates, the sensor's physical position can be determined experimentally using a rotating turntable and centripetal acceleration.

The measurement procedure utilizes the sensor logging application "PhyPhox":

1. Download and open "PhyPhox". Navigate to "Acceleration (without g)" and select the "Amount" (Betrag) tab.

2. Verify calibration by ensuring the displayed absolute acceleration reads approximately $0\,\text{m/s}^2$ while the smartphone lies flat and motionless on a level table.

3. Position the smartphone on the turntable aligned adjacent to its outer edge. Use a ruler to measure the distance $x$ from the central axis of rotation to the nearest edge of the phone casing. In this experimental setup, $x = 46\,\text{mm}$.

4. Begin data recording by pressing the play icon.

5. Engage the turntable at a preset rotation speed, selected here as $45\,\text{rpm}$ (alternative standard speeds include $33\,\text{rpm}$ and $78\,\text{rpm}$).

6. Once the rotation stabilizes and adequate data is gathered, stop the turntable, retrieve the smartphone, pause the recording, and review the acceleration plot.

Inspection of the graph during steady-state rotation yields a measured centripetal acceleration plateau of $a_z = 2.0\,\text{m/s}^2$.

The rotational frequency of the $45\,\text{rpm}$ setting is:

$f = \frac{45}{60\,\text{s}} = 0.75\,\text{Hz}$

The corresponding angular velocity is:

$\omega = 2\pi \cdot f = 2\pi \cdot 0.75\,\text{Hz} = 4.71\,\text{rad/s}$

Applying the centripetal acceleration formula gives the total distance $r$ from the turntable center to the internal accelerometer sensor:

$a_z = \omega^2 \cdot r \implies r = \frac{a_z}{\omega^2} = \frac{2.0\,\text{m/s}^2}{(4.71\,\text{rad/s})^2} = 0.090\,\text{m} = 90\,\text{mm}$

Subtracting the measured spacing $x$ gives the internal depth $d$ of the sensor relative to the smartphone edge:

$d = r - x = 90\,\text{mm} - 46\,\text{mm} = 44\,\text{mm}$

Repeating the procedure with the smartphone rotated by $90^\circ$ provides the perpendicular coordinate, thereby identifying the sensor's precise two-dimensional location on the device's internal circuit board.

Mechanics and Experimental Verification of Bicycle Cornering

When a cyclist navigates a curve, the bicycle must lean inward at an angle $\alpha$ relative to the vertical. Treating the bicycle and rider as a single lumped mass in uniform circular motion, stable cornering requires the net contact force from the ground to align along the axis connecting the tire contact patch to the system's center of mass. Any misalignment generates an unbalanced torque that causes the cyclist to tip over.

The force equilibrium is resolved into vertical and horizontal components:

Along the vertical (Y) axis, the upward normal force $F_N$ balances the downward gravitational force $F_g$:

$F_N = F_g \implies F_{\text{res},y} = 0$

Along the horizontal (X) axis, the inward static friction force $F_{HR}$ provides the centripetal acceleration:

$F_{HR} = F_{\text{res},x} = F_z$

The static friction force $F_{HR}$ adjusts to match the required centripetal demand; it is not automatically equal to the maximum static friction limit given by $\mu_{HR} \cdot F_N$.

From the vector geometry, the tangent of the lean angle $\alpha$ equals the ratio of the horizontal friction force to the vertical normal force:

$\tan(\alpha) = \frac{F_{HR}}{F_N} = \frac{F_z}{F_g} = \frac{\frac{m \cdot v^2}{r}}{m \cdot g} = \frac{v^2}{r \cdot g}$

Solving for the lean angle yields:

$\alpha = \arctan\left(\frac{v^2}{r \cdot g}\right)$

An outdoor experiment empirically validates this theoretical model using the following materials and procedures:

• Measuring tape (minimum length $8\,\text{m}$)

• Chalk

• Smartphone camera with high-speed burst mode

• Bicycle or electric scooter equipped with a digital speedometer

• Stopwatch or timing application

The protocol is conducted by student teams of 4 (with 5 being an exception):

1. Using chalk, a quarter-circle curve of radius $r = 8\,\text{m}$ is drawn on asphalt, leaving enough straight runway before and after for acceleration and deceleration. Within this circular arc, a dedicated segment of arc length $\Delta s = 5\,\text{m}$ is measured and marked.

2. One observer stands with a clear view of the $5\,\text{m}$ segment with a stopwatch. Another positions themselves facing the middle of the quarter circle with a camera to photograph the cyclist straight-on at the curve midpoint.

3. The coordinator signals when the course is clear. The cyclist accelerates to a target speed of approximately $12\,\text{km/h}$ ($v_1$) prior to entering the bend and maintains this speed steadily throughout the arc.

4. The timer records the elapsed transit time $\Delta t$ over the marked $5\,\text{m}$ path to calculate the true experimental velocity:

$v = \frac{\Delta s}{\Delta t}$

5. The photographer uses burst shooting to capture the cyclist at the exact midpoint, ensuring the lean angle $\alpha$ can be measured accurately against the vertical.

6. The run is repeated at higher target speeds of approximately $18\,\text{km/h}$ ($v_2$) and $24\,\text{km/h}$ ($v_3$).

7. The theoretical angles $\alpha_1, \alpha_2, \alpha_3$ are calculated from the measured speeds using the theoretical formula, and the actual angles are extracted from the photos using angle-measurement software or a triangle ruler.

8. The absolute angular error is evaluated across all three velocities:

$\Delta \alpha = |\alpha_{\text{calc}} - \alpha_{\text{meas}}|$

Experimental quality is classified based on angular deviation thresholds: an error $\Delta \alpha < 4^\circ$ indicates high measurement precision; an error between $4^\circ$ and $8^\circ$ represents acceptable accuracy with minor procedural errors; and an error $\Delta \alpha > 8^\circ$ indicates unacceptable discrepancies caused by speed variations, parallax, or improper angular evaluation.

Universal Gravitation

Newton's Law of Universal Gravitation states that any two bodies with masses $m_1$ and $m_2$ attract each other with a gravitational force $F_G$ directed along the line connecting their centers of mass:

$F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}$

The universal gravitational constant is established as:

$G = 6.6743 \times 10^{-11}\,\frac{\text{m}^3}{\text{kg}\cdot\text{s}^2} = 6.67 \times 10^{-11}\,\frac{\text{N}\cdot\text{m}^2}{\text{kg}^2}$

Although gravitational attraction operates between all matter over arbitrarily large distances, its intensity drops off with the square of the separation distance ($\propto 1/r^2$). Because $G$ is very small, gravitational forces between everyday objects are imperceptible and require massive bodies or sensitive instrumentation to detect.

Celestial Mechanics and Gravitational Phenomena

Gravitational theory enables the calculation of surface gravities, celestial masses, and laboratory interactions across diverse physical scales.

The Moon has an average radius of $r_M = 1738\,\text{km} = 1.738 \times 10^6\,\text{m}$ and a total mass of $m_M = 7.35 \times 10^{22}\,\text{kg}$. The surface gravitational acceleration $g_M$ is derived by equating gravitational force to weight:

$m \cdot g_M = G \cdot \frac{m \cdot m_M}{r_M^2} \implies g_M = \frac{G \cdot m_M}{r_M^2}$

$g_M = \frac{6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2 \cdot (7.35 \times 10^{22}\,\text{kg})}{(1.738 \times 10^6\,\text{m})^2} = 1.62\,\text{m/s}^2$

When astronaut Neil Armstrong jumped vertically on Earth ($g_E = 9.81\,\text{m/s}^2$), he reached an apex of $h_E = 0.60\,\text{m}$. Assuming identical takeoff velocity and mass, the initial kinetic energy is identical in both environments. At the peak of the jump, kinetic energy has converted entirely into potential energy ($E_{\text{pot}} = m \cdot g \cdot h$):

$m \cdot g_E \cdot h_E = m \cdot g_M \cdot h_M \implies h_M = \frac{h_E \cdot g_E}{g_M} = \frac{0.60\,\text{m} \cdot 9.81\,\text{m/s}^2}{1.62\,\text{m/s}^2} = 3.6\,\text{m}$

In 1787, Henry Cavendish first measured the weak gravitational attraction between laboratory masses using a torsion balance. Lead has a density of $\rho = 11.34 \times 10^3\,\text{kg/m}^3$.

To find the diameter $d$ of two identical lead spheres that, when touching ($r = d$), attract each other with a target force of $F_G = 0.010\,\text{N}$, the mass of each sphere is expressed as $m = \rho \cdot V = \rho \cdot \frac{4\pi}{3}\left(\frac{d}{2}\right)^3$:

$F_G = G \cdot \frac{m^2}{d^2} = G \cdot \frac{\rho^2 \cdot \left[\frac{4\pi}{3}\left(\frac{d}{2}\right)^3\right]^2}{d^2} = G \cdot \frac{\pi^2}{9} \cdot \rho^2 \cdot d^4$

Solving for the diameter:

$d = \sqrt[4]{\frac{36 \cdot F_G}{G \cdot \pi^2 \cdot \rho^2}} = \sqrt[4]{\frac{36 \cdot 0.010\,\text{N}}{6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2 \cdot \pi^2 \cdot (11.34 \times 10^3\,\text{kg/m}^3)^2}} = 1.4\,\text{m}$

The lead spheres used in Cavendish's actual experiment had diameters of $d_1 = 5.1\,\text{cm} = 0.051\,\text{m}$ and $d_2 = 20\,\text{cm} = 0.20\,\text{m}$. When placed in contact, their center-to-center distance is $r = \frac{d_1 + d_2}{2}$. The mutual gravitational force between them is:

$F_G = G \cdot \frac{m_1 \cdot m_2}{\left(\frac{d_1 + d_2}{2}\right)^2} = G \cdot \frac{\pi^2}{9} \cdot \frac{d_1^3 \cdot d_2^3}{(d_1 + d_2)^2} \cdot \rho^2$

$F_G = 6.67 \times 10^{-11} \cdot \frac{\pi^2}{9} \cdot \frac{(0.051\,\text{m})^3 \cdot (0.20\,\text{m})^3}{(0.051\,\text{m} + 0.20\,\text{m})^2} \cdot (11.34 \times 10^3\,\text{kg/m}^3)^2 = 1.6 \times 10^{-7}\,\text{N}$

Certain neutron stars rotate with a frequency of $f = 1.0\,\text{Hz}$. For material on the equatorial surface to remain gravitationally bound rather than being flung off by centrifugal effects, the inward gravitational attraction must equal or exceed the required centripetal force ($F_G \ge F_z$):

$G \cdot \frac{m \cdot M}{r^2} = m \cdot \frac{v^2}{r} = m \cdot \frac{(2\pi \cdot r \cdot f)^2}{r}$

Expressing the stellar mass as $M = \frac{4}{3}\pi \cdot r^3 \cdot \rho$:

$G \cdot \frac{\frac{4}{3}\pi \cdot r^3 \cdot \rho}{r^2} = 4\pi^2 \cdot r \cdot f^2$

The radius $r$ cancels completely out of the equation:

$G \cdot \frac{4}{3}\pi \cdot \rho = 4\pi^2 \cdot f^2 \implies \rho = \frac{3\pi \cdot f^2}{G}$

$\rho = \frac{3\pi \cdot (1.0\,\text{Hz})^2}{6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2} = 1.4 \times 10^{11}\,\text{kg/m}^3$

In May 1999, reports documented that the minor planet Ida (dimensions: length $l = 56 \times 10^3\,\text{m}$, width $b = 24 \times 10^3\,\text{m}$, height $h = 21 \times 10^3\,\text{m}$) is orbited by a small spherical moon named Dactyl (diameter $1.5\,\text{km}$) at an orbital radius of $r = 100\,\text{km} = 100 \times 10^3\,\text{m}$ with an orbital period of $T = 22\,\text{h} = 79200\,\text{s}$.

Equating gravitational force to centripetal force yields the mass of Ida:

$G \cdot \frac{m \cdot M}{r^2} = m \cdot \frac{v^2}{r} = m \cdot \frac{\left(\frac{2\pi \cdot r}{T}\right)^2}{r} \implies M = \frac{4\pi^2 \cdot r^3}{G \cdot T^2}$

Approximating Ida's volume as a rectangular cuboid ($V = l \cdot b \cdot h$), its density is calculated as:

$\rho = \frac{M}{V} = \frac{4\pi^2 \cdot r^3}{G \cdot T^2 \cdot l \cdot b \cdot h}$

$\rho = \frac{4\pi^2 \cdot (100 \times 10^3\,\text{m})^3}{6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2 \cdot (79200\,\text{s})^2 \cdot (56 \times 10^3\,\text{m}) \cdot (24 \times 10^3\,\text{m}) \cdot (21 \times 10^3\,\text{m})} = 3.3 \times 10^3\,\text{kg/m}^3$

This confirms that Ida has a density similar to that of the Moon, consisting predominantly of silicate rock.

At the center of the Milky Way, a supermassive black hole is orbited by a star with an observed period of $T = 15\,\text{a} = 15 \times 365.25 \times 86400\,\text{s} = 4.7 \times 10^8\,\text{s}$ at a mean orbital radius of $r = 0.013\,\text{LJ} = 1.2 \times 10^{14}\,\text{m}$.

Balancing gravitational force and centripetal force:

$G \cdot \frac{m \cdot M}{r^2} = m \cdot \frac{v^2}{r} \implies \frac{G \cdot M}{r} = \left(\frac{2\pi \cdot r}{T}\right)^2$

$M = \frac{4\pi^2 \cdot r^3}{T^2 \cdot G} = \frac{4\pi^2 \cdot (1.2 \times 10^{14}\,\text{m})^3}{(4.7 \times 10^8\,\text{s})^2 \cdot 6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2} = 4.9 \times 10^{36}\,\text{kg}$

Comparing this mass to the solar mass ($M_S = 1.9884 \times 10^{30}\,\text{kg}$):

$\frac{M}{M_S} = \frac{4.9 \times 10^{36}\,\text{kg}}{1.9884 \times 10^{30}\,\text{kg}} = 2.5 \times 10^6$

The supermassive black hole contains approximately $2.5 \times 10^6$ solar masses.