The variables r and are shown in Figure 13.17 in the case of an ellipse. Continue reading with a Scientific American subscription. So in this type of case, scientists use the spacecrafts orbital period near the planet or any other passing by objects to determine the planets gravitational pull. Newton, building on other people's observations, showed that the force between two objects is proportional to the product of their masses and decreases with the square of the distance: where \(G=6.67 \times 10^{-11}\) m\(^3\)kg s\(^2\) is the gravitational constant. are licensed under a, Coordinate Systems and Components of a Vector, Position, Displacement, and Average Velocity, Finding Velocity and Displacement from Acceleration, Relative Motion in One and Two Dimensions, Potential Energy and Conservation of Energy, Rotation with Constant Angular Acceleration, Relating Angular and Translational Quantities, Moment of Inertia and Rotational Kinetic Energy, Gravitational Potential Energy and Total Energy, Comparing Simple Harmonic Motion and Circular Motion, (a) An ellipse is a curve in which the sum of the distances from a point on the curve to two foci, As before, the distance between the planet and the Sun is. The gravitational attraction between the Earth and the sun is G times the sun's mass times the Earth's mass, divided by the distance between the Earth and the sun squared. then you must include on every digital page view the following attribution: Use the information below to generate a citation. It's a matter of algebra to tease out the mass by rearranging the equation to solve for M . This is a direct application of Equation \ref{eq20}. In practice, the finite acceleration is short enough that the difference is not a significant consideration.) The prevailing view during the time of Kepler was that all planetary orbits were circular. Because the value of and G is constant and known. For any ellipse, the semi-major axis is defined as one-half the sum of the perihelion and the aphelion. one or more moons orbitting around a double planet system. Equation 13.8 gives us the period of a circular orbit of radius r about Earth: For an ellipse, recall that the semi-major axis is one-half the sum of the perihelion and the aphelion. Jan 19, 2023 OpenStax. squared cubed divided by squared can be used to calculate the mass, , of a universal gravitation using the sun's mass. How to decrease satellite's orbital radius? have moons, they do exert a small pull on one another, and on the other planets of the solar system. Take for example Mars orbiting the Sun. Kepler's 3rd law can also be used to determine the fast path (orbit) from one planet to another. Mars is closest to the Sun at Perihelion and farthest away at Aphelion. 9 / = 1 7 9 0 0 /. Following on this observations Kepler also observed the orbital periods and orbital radius for several planets. The mass of Earth is 598 x 1022 kg, which is 5,980,000,000,000,000,000,000,000 kg (598 with 22 zeros after that). 1017 0 obj <>stream These last two paths represent unbounded orbits, where m passes by M once and only once. Kepler's Third Law - average radius instead of semimajor axis? A planet is discovered orbiting a The most efficient method is a very quick acceleration along the circular orbital path, which is also along the path of the ellipse at that point. Now, we have been given values for And those objects may be any, a moon orbiting the planet with a mass of, the distance between the moon and the planet is, To maintain the orbital path, the moon would also act, Where T is the orbital period of the moon around that planet. Is this consistent with our results for Halleys comet? It is impossible to determine the mass of any astronomical object. Copyright 2023 NagwaAll Rights Reserved. Mar 18, 2017 at 3:12 Your answer is off by about 31.5 Earth masses because you used a system that approximates this system. squared times 9.072 times 10 to the six seconds quantity squared. F= ma accel. have the sun's mass, we can similarly determine the mass of any planet by astronomically determining the planet's orbital Well, suppose we want to launch a satellite into outer space that will orbit the Earth at a specified orbital radius, \(R_s\). Best!! A note about units: you should use what units make sense as long as they are consistent, ie., they are the same for both of the orbital periods and both orbital radii, so they cancel out. Say that you want to calculate the centripetal acceleration of the moon around the Earth. The ratio of the periods squared of any two planets around the sun is equal to the ratio of their average distances from the sun cubed. Find the orbital speed. Nothing to it. equals four squared cubed We are know the orbital period of the moon is \(T_m = 27.3217\) days and the orbital radius of the moon is \(R_m = 60\times R_e\) where \(R_e\) is the radius of the Earth. An ellipse is defined as the set of all points such that the sum of the distance from each point to two foci is a constant. The green arrow is velocity. has its path bent by an amount controlled by the mass of the asteroid. By observing the time between transits, we know the orbital period. [closed], Improving the copy in the close modal and post notices - 2023 edition, New blog post from our CEO Prashanth: Community is the future of AI, Calculating specific orbital energy, semi-major axis, and orbital period of an orbiting body. See Answer Answer: T planet . For each planet he considered various relationships between these two parameters to determine how they were related. Whereas, with the help of NASAs spacecraft. See the NASA Planetary Fact Sheet, for fundamental planetary data for all the planets, and some moons in our solar system. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. of distant astronomical objects (Exoplanets) is determined by the objects apparent size and shape. The purple arrow directed towards the Sun is the acceleration. Physics . And while the astronomical unit is Kepler's Third Law can also be used to study distant solar systems. And those objects may be any moon (natural satellite), nearby passing spacecraft, or any other object passing near it. Can corresponding author withdraw a paper after it has accepted without permission/acceptance of first author. Help others and share. Figure 13.19 shows the case for a trip from Earths orbit to that of Mars. Substituting, \[\begin{align*} \left(\frac{T_s}{T_m}\right)^2 &=\left(\frac{R_s}{R_m}\right)^3 \\[4pt] T_s^2 &=T_m^2\left(\frac{R_s}{R_m}\right)^3 \\[4pt] T_s &=T_m\left(\frac{R_s}{R_m}\right)^{\frac{3}{2}} \\[4pt] &=27.3217\left(\frac{6 R_e}{60 R_e}\right)^{\frac{3}{2}} \\[4pt] &=27.3217\left(\frac{1 }{10 }\right)^{\frac{3}{2}} \\[4pt] &=27.3217\left(0.0317\right) \\[4pt] &= 0.86\;days \end{align*}\]. The Attempt at a Solution 1. To do that, I just used the F=ma equation, with F being the force of gravity, m being the mass of the planet, and a =v^2/r. orbit around a star. Which language's style guidelines should be used when writing code that is supposed to be called from another language? The time it takes a planet to move from position A to B, sweeping out area A1A1, is exactly the time taken to move from position C to D, sweeping area A2A2, and to move from E to F, sweeping out area A3A3. In Satellite Orbits and Energy, we derived Keplers third law for the special case of a circular orbit. JavaScript is disabled. Consider Figure 13.20. For the Hohmann Transfer orbit, we need to be more explicit about treating the orbits as elliptical. are not subject to the Creative Commons license and may not be reproduced without the prior and express written The other important thing to note, is that it is not very often that the orbits line up exactly such that a Hohmann transfer orbit is possible. How do I calculate a planet's mass given a satellite's orbital period and semimajor axis? Hence from the above equation, we only need distance between the planet and the moon r and the orbital period of the moon T to calculate the mass of a planet. Consider a planet with mass M planet to orbit in nearly circular motion about the sun of mass . Consider using vis viva equation as applied to circular orbits. we have equals four squared times 7.200 times 10 to the 10 meters quantity I know the solution, I don't know how to get there. So if we can measure the gravitational pull or acceleration due to the gravity of any planet, we can measure the mass of the planet. This is information outside of the parameters of the problem. We and our partners use data for Personalised ads and content, ad and content measurement, audience insights and product development. To move onto the transfer ellipse from Earths orbit, we will need to increase our kinetic energy, that is, we need a velocity boost. How do I figure this out? \frac{T^2_{Moon}}{T^2_s}=19^2\sim 350 Answer. Hence, to travel from one circular orbit of radius r1r1 to another circular orbit of radius r2r2, the aphelion of the transfer ellipse will be equal to the value of the larger orbit, while the perihelion will be the smaller orbit. x~\sim (19)^2\sim350, For example, NASAs space probes Voyager 1 and Voyager 2 were used to measuring the outer planets mass. This "bending" is measured by careful tracking and Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. There are other options that provide for a faster transit, including a gravity assist flyby of Venus. M_p T^2_s\approx M_{Earth} T^2_{Moon}\quad \Rightarrow\quad \frac{M_p}{M_{Earth}}\approx centripetal force is the Earth's mass times the square of its speed divided by its distance from the sun. INSTRUCTIONS: Choose units and enter the following: Planetary Mass (M): The calculator returns the mass (M) in kilograms. YMxu\XQQ) o7+'ZVsxWfaEtY/ vffU:ZI k{z"iiR{5( jW,oxky&99Wq(k^^YY%'L@&d]a K Therefore we can set these two forces equal, \[ \frac{GMm}{r^2} =\frac{mv^2}{r} \nonumber\]. Kepler's Third Law. Because we know the radius of the Earth, we can use the Law of Universal Gravitation to calculate the mass of the Earth in terms of the decimal places, we have found that the mass of the star is 2.68 times 10 to the 30 4. follow paths that are subtly different than they would be without this perturbing effect. Now, let's consider the fastest path from Earth to Mars using Kepler's Third Law. 1024 kg. However, it seems (from the fact that the object is described as being "at rest") that your exercise is not assuming an inertial reference frame, but rather a rotating reference frame matching the rotation of the planet. then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a digital format, Consider two planets (1 and 2) orbiting the sun. As a result, the planets determining the distance to the sun, we can calculate the earth's speed around the sun and hence the sun's mass. An example of data being processed may be a unique identifier stored in a cookie. group the units over here, making sure to distribute the proper exponents. satellite orbit period: satellite mean orbital radius: planet mass: . L=rp=r(prad+pperp)=rprad+rpperpL=rp=r(prad+pperp)=rprad+rpperp. Use Kepler's law of harmonies to predict the orbital period of such a planet. , which is equal to 105 days, and days is not the SI unit of time. I should be getting a mass about the size of Jupiter. Next, well look at orbital period, %PDF-1.5 % But few planets like Mercury and Venus do not have any moons. Visit this site for more details about planning a trip to Mars. This yields a value of 2.671012m2.671012m or 17.8 AU for the semi-major axis. And thus, we have found that Newton's Law of Gravitation states that every bit of matter in the universe attracts every other . @ZeroTheHero: I believe the Earth-Sun distance is about 8 light-minutes, I guess it's the Earth-Moon distance that is about 1 light-second, but then, it seems, the mass of the planet is much smaller than that of the Earth. You can also view the more complicated multiple body problems as well. %%EOF We recommend using a The first term on the right is zero because rr is parallel to pradprad, and in the second term rr is perpendicular to pperppperp, so the magnitude of the cross product reduces to L=rpperp=rmvperpL=rpperp=rmvperp. The velocity boost required is simply the difference between the circular orbit velocity and the elliptical orbit velocity at each point. Apparently I can't just plug these in to calculate the planets mass. used frequently throughout astronomy, its not in SI unit. \frac{M_p}{M_E}=\frac{a_s^3T_M^2}{a_M^3 T_s^2}\, . And returning requires correct timing as well. (You can figure this out without doing any additional calculations.) This book uses the The constant of proportionality depends on the mass, \(M\) of the object being orbited and the gravitational constant, \(G\). It turned out to be considerably lighter and more "frothy" in structure than had been expected, a fact I attempted to find the velocity from the radius (2.6*10^5) and the time (2.5hr*60*60=9000s) The same (blue) area is swept out in a fixed time period. For planets without observable natural satellites, we must be more clever. xYnF}Gh7\.S !m9VRTh+ng/,4sY~TfeAe~[zqqR f2}>(c6PXbN%-o(RgH_4% CjA%=n o8!uwX]9N=vH{'n^%_u}A-tf>4\n In such a reference frame the object lying on the planet's surface is not following a circular trajectory, but rather appears to be motionless with respect to the frame of . First, we have not accounted for the gravitational potential energy due to Earth and Mars, or the mechanics of landing on Mars. Conversions: gravitational acceleration (a) Keplers third law states that the square of the period is proportional to the cube of the semi-major axis of the orbit. For a circular orbit, the semi-major axis (a) is the same as the radius for the orbit. Accessibility StatementFor more information contact us atinfo@libretexts.org. In fact, because almost no planet, satellite, or moon is actually on a perfectly circular orbit \(R\) is the semi-major axis of the elliptical path of the orbiting object. escape or critical speed: planet mass: planet radius: References - Books: Tipler, Paul A.. 1995. The consent submitted will only be used for data processing originating from this website. This page titled 3.1: Orbital Mechanics is shared under a CC BY-SA license and was authored, remixed, and/or curated by Magali Billen. Planetary scientists also send orbiters to other planets to make similar measurements (okay not vegetation). several asteroids have been (or soon will be) visited by spacecraft. calculate. Manage Settings (In fact, the acceleration should be instantaneous, such that the circular and elliptical orbits are congruent during the acceleration. I see none of that being necessary here, it seems to me that it should be solvable using Kepler's Laws although I may be wrong about that. For the case of orbiting motion, LL is the angular momentum of the planet about the Sun, rr is the position vector of the planet measured from the Sun, and p=mvp=mv is the instantaneous linear momentum at any point in the orbit. Although the mathematics is a bit For the case of traveling between two circular orbits, the transfer is along a transfer ellipse that perfectly intercepts those orbits at the aphelion and perihelion of the ellipse. We can double . I attempted to use Kepler's 3rd Law, Note that when the satellite leaves the Earth, Mars will not yet be at Perihelion, so the timing is important. If a satellite requires 2.5 h to orbit a planet with an orbital radius of 2.6 x 10^5 m, what is the mass of the planet? The values of and e determine which of the four conic sections represents the path of the satellite. hbbd``b`$W0H0 # ] $4A*@+Hx uDB#s!H'@ % This is the how planetary scientists determined the mass of Earth, the mass of other planets in our solar system that have moons, the mass of the moon using an orbiter, and the mass of other stars when orbiting planets can be observed. So scientists use this method to determine the planets mass or any other planet-like objects mass. seconds. The Planet's Mass from Acceleration and Radius calculator computes the mass of planet or moon based on the radius (r), acceleration due to gravity on the surface (a) and the universal gravitational constant (G). Recall that a satellite with zero total energy has exactly the escape velocity. The time taken by an object to orbit any planet depends on that. Where does the version of Hamapil that is different from the Gemara come from? If there are any complete answers, please flag them for moderator attention. We must leave Earth at precisely the correct time such that Mars will be at the aphelion of our transfer ellipse just as we arrive. We can use these three equalities Substituting them in the formula, 994 0 obj <> endobj OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. We also need the Constant of Proportionality in the Law of Universal Gravitation, G. This value was experimentally determined Homework Statement What is the mass of a planet (in kg and in percent of the mass of the sun), if: its period is 3.09 days, the radius of the circular orbit is 6.43E9 m, and the orbital velocity is 151 km/s. $$ However, knowing that it is the fastest path places clear limits on missions to Mars (and similarly missions to other planets) including sending manned missions. We conveniently place the origin in the center of Pluto so that its location is xP=0. \frac{M_pT_s^2}{a_s^3}=\frac{M_E T_M^2}{a_M^3} \quad \Rightarrow \quad Instead I get a mass of 6340 suns. measurably perturb the orbits of the other planets? So the order of the planets in our solar system according to mass is, NASA Mars Perseverance Rover {Facts and Information}, Haumea Dwarf Planet Facts and Information, Orbit of the International Space Station (ISS), Exploring the Number of Planets in Our Solar System and Beyond, How long is a day and year on each planet, Closest and farthest distance of each planet, How big are the stars? hours, and minutes, leaving only seconds. Although Mercury and Venus (for example) do not (T is known), Hence from the above equation, we only need distance between the planet and the moon r and the orbital period of the moon T, So scientists use this method to determine the, Now as we knew how to measure the planets mass, scientists used their moons for planets like, Space probes are one of the ways for determining the gravitational pull and hence the mass of a planet. understanding of physics and some fairly basic math, we can use information about a Additional detail: My class is working on velocity and acceleration in polar coordinates with vectors. Recently, the NEAR spacecraft flew by the asteroid Mathilde, determining for the Lets take a closer look at the Give your answer in scientific You can see an animation of two interacting objects at the My Solar System page at Phet. By Jimmy Raymond \[ \left(\frac{2\pi r}{T}\right)^2 =\frac{GM}{r} \]. If the planet in question has a moon (a natural satellite), then nature has already done the work for us. Space probes are one of the ways for determining the gravitational pull and hence the mass of a planet. What is the mass of the star? For a circular orbit, the semi-major axis ( a) is the same as the radius for the orbit. By observing the time it takes for the satellite to orbit its primary planet, we can utilize Newton's equations to infer what the mass of the planet must be. As before, the Sun is at the focus of the ellipse. It may not display this or other websites correctly. You can view an animated version of Figure 13.20, and many other interesting animations as well, at the School of Physics (University of New South Wales) site. Knowing the mass and radius of the Earth and the distance of the Earth from the sun, we can calculate the mass of the For curiosity's sake, use the known value of g (9.8 m/s2) and your average period time, and . possible period, given your uncertainties. You may find the actual path of the Moon quite surprising, yet is obeying Newtons simple laws of motion. so lets make sure that theyre all working out to reach a final mass value in units Want to cite, share, or modify this book? Additional details are provided by Gregory A. Lyzenga, a physicist at Harvey Mudd College in Claremont, Calif. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo But another problem was that I needed to find the mass of the star, not the planet. The angle between the radial direction and v v is . Learn more about Stack Overflow the company, and our products. In astronomy, planetary mass is a measure of the mass of a planet-like astronomical object.Within the Solar System, planets are usually measured in the astronomical system of units, where the unit of mass is the solar mass (M ), the mass of the Sun.In the study of extrasolar planets, the unit of measure is typically the mass of Jupiter (M J) for large gas giant planets, and the mass of . Saturn Distance from Sun How Far is Planet Saturn? Using a telescope, one can detect other planets around stars by observing a drop in the brightness of the star as the planet transits between the star and the telescope. I have a homework question asking me to calculate the mass of a planet given the semimajor axis and orbital period of its moon. Nagwa is an educational technology startup aiming to help teachers teach and students learn. But these other options come with an additional cost in energy and danger to the astronauts. Rearranging the equation gives: M + m = 42r3 GT 2. Substituting for the values, we found for the semi-major axis and the value given for the perihelion, we find the value of the aphelion to be 35.0 AU. Continue with Recommended Cookies. And now multiplying through 105 first time its actual mass. The last step is to recognize that the acceleration of the orbiting object is due to gravity. These are the two main pieces of information scientists use to measure the mass of a planet. However, this can be automatically converted into other mass units via the pull-down menu including the following: This calculator computes the mass of a planet given the acceleration at the surface and the radius of the planet. With this information, model of the planets can be made to determine if they might be convecting like Earth, and if they might have plate tectonics. In practice, that must be part of the calculations. meaning your planet is about $350$ Earth masses. All motion caused by an inverse square force is one of the four conic sections and is determined by the energy and direction of the moving body. We know that the path is an elliptical orbit around the sun, and it grazes the orbit of Mars at aphelion. In order to use gravity to find the mass of a planet, we must somehow measure the strength of its "tug" on another object. Planets in Order from Smallest to Largest. This behavior is completely consistent with our conservation equation, Equation 13.5. So, without ever touching a star, astronomers use mathematics and known physical laws to figure out its mass. Now, lets cancel units of meters You do not want to arrive at the orbit of Mars to find out it isnt there. How do we know the mass of the planets? The ratio of the dimensions of the two paths is the inverse of the ratio of their masses. You can also use orbital velocity and work it out from there. The equation for centripetal acceleration means that you can find the centripetal acceleration needed to keep an object moving in a circle given the circle's radius and the object's angular velocity. Learn more about our Privacy Policy. What is the mass of the star? The areal velocity is simply the rate of change of area with time, so we have. $3.8\times 10^8$ is barely more than one light-second, which is about the Earth-Sun distance, but the orbital period of the Moon is about 28 days, so you need quite a bit of mass ($\sim 350$ Earth masses?) planet or star given the orbital period, , and orbital radius, , of an object Calculate the lowest value for the acceleration. That is, for each planet orbiting another (much larger) object (the Sun), the square of the orbital period is proportional to the cube of the orbital radius. 1.5 times 10 to the 11 meters. Why the obscure but specific description of Jane Doe II in the original complaint for Westenbroek v. Kappa Kappa Gamma Fraternity? But first, let's see how one can use Kepler's third law to for two applications. We can resolve the linear momentum into two components: a radial component pradprad along the line to the Sun, and a component pperppperp perpendicular to rr. Is "I didn't think it was serious" usually a good defence against "duty to rescue"? Note the mass of Jupiter is ~320 times the mass of Earth, so you have a Jupiter-sized planet. As with Keplers first law, Newton showed it was a natural consequence of his law of gravitation. Discover world-changing science. If the total energy is exactly zero, then e=1e=1 and the path is a parabola. the orbital period and the density of the two objectsD.) The semi-major axis, denoted a, is therefore given by a=12(r1+r2)a=12(r1+r2). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Kepler's third law provides an accurate description of the period and distance for a planet's orbits about the sun. Since the object is experiencing an acceleration, then there must also be a force on the object. citation tool such as, Authors: William Moebs, Samuel J. Ling, Jeff Sanny. Based on measurements of a moon's orbit with respect to the planet, what can one calculate? For the moment, we ignore the planets and assume we are alone in Earths orbit and wish to move to Mars orbit. The constant e is called the eccentricity. Scientists also measure one planets mass by determining the gravitational pull of other planets on it. There are four different conic sections, all given by the equation. This path is the Hohmann Transfer Orbit and is the shortest (in time) path between the two planets. Kepler's Third law can be used to determine the orbital radius of the planet if the mass of the orbiting star is known (\(R^3 = T^2 - M_{star}/M_{sun} \), the radius is in AU and the period is in earth years). planet mass: radius from the planet center: escape or critical speed. A small triangular area AA is swept out in time tt. Sometimes the approximate mass of distant astronomical objects (Exoplanets) is determined by the objects apparent size and shape. T 2 = 42 G(M + m) r3. cubed divided by 6.67 times 10 to the negative 11 meters cubed per kilogram second Now as we knew how to measure the planets mass, scientists used their moons for planets like Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Dwarf Planet Pluto, and objects those have moons. Newton's second Law states that without such an acceleration the object would simple continue in a straight line. It is labeled point A in Figure 13.16. , the universal gravitational The mass of all planets in our solar system is given below. I have a semimajor axis of $3.8\times10^8$ meters and a period of $1.512$ days. This answer uses the Earth's mass as well as the period of the moon (Earth's moon). In the above discussion of Kepler's Law we referred to \(R\) as the orbital radius. We can find the circular orbital velocities from Equation 13.7. Thanks for reading Scientific American. Whereas, with the help of NASAs spacecraft MESSENGER, scientists determined the mass of the planet mercury accurately. Issac Newton's Law of Universal Gravitation tells us that the force of attraction between two objects is proportional the product of their masses divided by the square of the distance between their centers of mass. We now have calculated the combined mass of the planet and the moon. A boy can regenerate, so demons eat him for years. Many geological and geophysical observations are made with orbiting satellites, including missions that measure Earth's gravity field, topography, changes in topography related to earthquakes and volcanoes (and other things), and the magnetic field. The Mass of a planet The mass of the planets in our solar system is given in the table below. The Sun is not located at the center of the ellipse, but slightly to one side (at one of the two foci of the ellipse). So we have some planet in circular Lets take the case of traveling from Earth to Mars. Which reverse polarity protection is better and why? Orbital radius and orbital period data for the four biggest moons of Jupiter are listed in the . Hence we find He determined that there is a constant relationship for all the planets orbiting the sun. gravitational force on an object (its weight) at the Earth's surface, using the radius of the Earth as the distance. Since the angular momentum is constant, the areal velocity must also be constant. Start with the old equation Doppler radio measurement from Earth. $$ For an ellipse, recall that the semi-major axis is one-half the sum of the perihelion and the aphelion. Now, we calculate \(K\), \[ \begin{align*} K&=\frac{4\pi^2}{GM} \\[4pt] &=2.97 \times 10^{-19}\frac{s^2}{m^3} \end{align*}\], For any object orbiting the sun, \(T^2/R^3 = 2.97 \times 10^{-19} \), Also note, that if \(R\) is in AU (astonomical units, 1 AU=1.49x1011 m) and \(T\) is in earth-years, then, Now knowing this proportionality constant is related to the mass of the object being orbited, gives us the means to determine the mass this object by observing the orbiting objects.