PHY 101 Lecture 6: Kepler's Laws & Gravitational Potential PDF

# Chapter F ## Introduction to Kepler's Laws of Planetary Motion When we talk about Kepler’s law of planetary motion, the concept of “motion” is always relative. Here, the motion is divided into two categories: - Bounded Motion - Unbounded Motion ### Bounded Motion In bounded motion, the particle...

# Chapter F ## Introduction to Kepler's Laws of Planetary Motion When we talk about Kepler’s law of planetary motion, the concept of “motion” is always relative. Here, the motion is divided into two categories: - Bounded Motion - Unbounded Motion ### Bounded Motion In bounded motion, the particle has negative total energy (E < 0) and has two or more extreme points where the total energy is always equal to the potential energy of the particle, i.e., the kinetic energy of the particle becomes zero. For eccentricity 0 ≤ e < 1, E < 0 implies the body has bounded motion. A circular orbit has eccentricity e = 0, and an elliptical orbit has eccentricity e < 1. ### Unbounded Motion In unbounded motion, the particle has positive total energy (E > 0) and has a single extreme point where the total energy is always equal to the potential energy of the particle, i.e., the kinetic energy of the particle becomes zero. For eccentricity e ≥ 1, E > 0 implies the body has unbounded motion. The parabolic orbit has eccentricity e = 1, and the hyperbolic path has eccentricity e > 1. Kepler's laws of planetary motion can be stated as follows: ## Kepler's First Law According to Kepler's first law, "All the planets revolve around the sun in elliptical orbits having the sun at one of the foci". The point at which the planet is close to the sun is known as **perihelion** (about 147 million kilometers from the sun), and the point at which the planet is farther from the sun is known as **aphelion** (152 million kilometers from the sun). It is characteristic of an ellipse that the sum of the distances of any planet from two foci is constant. ## Kepler's Second Law Kepler's second law states, “The radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time”. As the orbit is not circular, the planet’s kinetic energy is not constant in its path. It has more kinetic energy near the perihelion, and less kinetic energy near the aphelion implies more speed at the perihelion and less speed (*v_min*) at the aphelion. If *r* is the distance of the planet from the sun at perihelion (*r_min*) and at aphelion (*r_max*), then, $ \frac{r_{min}+r_{max}}{2}= 2a$, where a is the length of the major axis of an ellipse… (1) ## Kepler's Third Law According to Kepler's law of periods, “The square of the period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis”. $T^2 \propto r^3$ The shorter the orbit of the planet around the sun, the shorter the time taken to complete one revolution. Using the equations of Newton's law of gravitation and laws of motion, Kepler's third law takes a more general form. $P² = \frac{4\pi²}{G (Ma₁+Ma₂)}xr³$ Where Ma₁ and Ma₂ are the masses of the two orbiting objects in solar masses. ## Gravitational Potential Energy When a body of mass (*m*) is moved from infinity to a point inside the gravitational influence of a source mass (*M*) without accelerating it, the amount of work done in displacing it into the source field is stored in the form of potential energy. This is known as gravitational potential energy. It is represented by the symbol *U_g*. Potential energy of a body at a given position is defined as the energy stored in the body at that position. If the position of the body changes due to the application of external forces, the change in potential energy is equal to the amount of work done on the body by the forces. Under the action of gravitational force, the work done is independent of the path taken for a change in position, so the force is a conservative force. Besides, all such forces have some potential in them. The gravitational influence on a body at infinity is zero; therefore, potential energy is zero, which is called a reference point. The gravitational potential energy is: $GPE = mgh$, where: - *m* is the mass in kilograms - *g* is the acceleration due to gravity (9.8 on earth) - *h* is the height above the ground in metres ### Derivation of Gravitational Potential Energy Consider a source mass 'M' is placed at a point along the x-axis; initially, at rest mass 'm' is at infinity. A small amount of work done in bringing it without acceleration through a very small distance (dx) which is given by $dw = Fdx$ Here, F is an attractive force, and the displacement is towards the negative x-axis direction, so F and dx are in the same direction. Then, $dw = \{\frac{GMm}{x²}\}dx$ Integrating on both sides $W = \int_{\infty}^{r} \frac{GMm}{x²}dx$ $W = GMm \{\frac{1}{x}\}_{\infty}^{r}$ $W = \frac{GMm}{r} -(\frac{GMm}{\infty})$ $W = \frac{GMm}{r}$ Since the work done is stored as its potential energy *U*, the gravitational potential energy at a point which is at a distance 'r' from the source mass is given by; $U = \frac{-GMm}{r}$ If a test mass moves from a point inside the gravitational field to the other point inside the same gravitational field of source mass, then the change in potential energy of the test mass is given by; $\Delta U = GMm \{\frac{1}{r_{i}}-\frac{1}{r_{r}}\}$ If $r_{i} > r_{r}$ then $\Delta U$ is negative. ### Expression for Gravitational Potential Energy at Height (h) - Derive $\Delta U = mgh$ If a body is taken from the surface of the earth to a point at a height 'h' above the surface of the earth, then $r_{i} = R$ and $r_{r} = R + h$, then, $\Delta U = GMm \{\frac{1}{R} - \frac{1}{(R+h)}\}$ $\Delta U = \frac{GMmh}{R(R+h)}$ When $h << R$, then $R + h = R$ and $g = \frac{GM}{R²}$ On substituting this in the above equation, we get, Gravitational Potential Energy $\Delta U = mgh$ ## Gravitational Potential The amount of work done in moving a unit test mass from infinity into the gravitational influence of source mass is known as gravitational potential. Simply, it is the gravitational potential energy possessed by a unit test mass. $V = \frac{U}{m}$ $V = \frac{GM}{r}$ ### Relation between Gravitational Field Intensity and Gravitational Potential #### Integral Form: $V = -\Sigma E.dr$ (If E is given then v has to be found using this formula) #### Differential Form: $E = -\frac{dv}{dr}$ (If Vis given then Ehas to be found using this formula) $E = \frac{\partial v}{\partial x} i + \frac{\partial v}{\partial y} j + \frac{\partial v}{\partial z} k$ (components along x, y and z directions). ## Gravitational Potential of a Point Mass Consider a point mass M, the gravitational potential at a distance 'r' from it is given by; $ V = \frac{GM}{r}$ ## Gravitational Potential of a Spherical Shell Consider a thin uniform spherical shell of the radius (R) and mass (M) situated in space. Now, - **Event 1: If point 'P' lies inside the spherical shell (r< R):** As $E = 0$, *V* is a constant. The value of gravitational potential is given by, *V = -GM/R*. - **Event 2: If point 'P' lies on the surface of the spherical shell (r=R):** On the surface of the earth, $E = -GM/R²$. Using the relation $V = -\Sigma E.dr$ over a limit of (0 to R), we get, Gravitational Potential (*V*) = -GM/R. - **event 3: If point 'P' lies outside the spherical shell (r>R):** Outside the spherical shell, $E = -GM/r²$. Using the relation: $V = -\Sigma E.dr$ over a limit of (0 to r), then, $V = -GM/r$. ## Gravitational Potential of a Uniform Solid Sphere Consider a thin, uniform solid sphere of radius (R) and mass (M) situated in space. Now, - **Event 1: If point 'P' lies inside the uniform solid sphere (r< R):** Inside the uniform solid sphere, $E = -GMr/R³$. Using the relation $V = -\Sigma E.dr$ over a limit of (0 to r). The value of gravitational potential is given by, $v = -GM\{\frac{(3R^2-r^2)}{2R^2}\}$ - **Event 2: If point 'P' lies on the surface of the uniform solid sphere (r = R):** On the surface of a uniform solid sphere, $E = -GM/R²$. Using the relation $V = -\Sigma E.dr$ over a limit of (0 to R) we get, $V = \frac{GM}{R}$. - **Event 3: If point 'P' lies outside the uniform solid sphere (r> R):** Using the relation over a limit of (0 to r), we get, $V = \frac{GM}{r}$. - **Event 4: Gravitational potential at the centre of the solid sphere is given by** $V = -\frac{3}{2} \frac{GM}{R}$ ## Escape and Orbital Velocity In order to define escape and orbital velocities in simple words, we can say that orbital velocity is the speed required by an object to orbit around another object, whereas escape velocity is the minimum speed required by an object to break free from the gravitational pull of another mass. The concepts of escape and orbital velocities and their relationship are discussed below. ### Escape Velocity The studied proof that in kinematics the range of a projectile depends upon the initial velocity as $R_{max} \propto u^2$, $R_{max} = \frac{u^2}{2g}$, which implies that for a certain initial velocity provided to the particle, the particle flies away from the gravitational influence of the Earth. This minimum amount of velocity for which the particle escapes the gravitational sphere of influence of a planet is known as escape velocity (*V_e*). When escape velocity is provided to a body, it goes to infinity theoretically. As the gravitational force is a conservative force, the law of conservation of energy holds good. Applying the law of conservation of energy for the particle which is provided with a required minimum velocity to go to infinity, $U_i + K_i = U_f + K_f$ At infinity, the particles experience no interaction so the final potential energy and final velocity of a body becomes zero after reaching its maximum height. So from this, it can deduce the final kinetic energy of the particle. Hence, $U_f + K_f = 0$, we then have $U_i = \frac{-GMm}{R}$, $K_i = \frac{1}{2}mv^2$ From the equation we get, $\frac{1}{2}mv^2 + (\frac{-GMm}{R} ) = 0$, $ \frac{1}{2}mv^2 = \frac{GMm}{R}$ $v = \sqrt{\frac{2GM}{R}}$………….(1) From the above formula, it is clear that escape velocity does not depend upon the mass (*m*). If the source mass is earth, then the escape velocity has a value of 11.2 km/s. If $v = v_e$, then the body escapes the planet’s gravitational sphere of influence; if $0 ≤ v ≤ v_e$, then the body either falls back onto the earth or continues to orbit around the planet within the sphere of influence of the planet. ### Orbital Velocity The velocity with which the test mass orbits around a source mass is known as orbital velocity (*V_o*). When the test mass is orbiting around the source mass in a circular path of radius 'r' having a centre of the source mass as the centre of the circular path, the centripetal force is provided by the gravitational force, as it is always an attracting force having its direction pointed towards the centre of a source mass. $\frac{mv^2}{r} = \frac{GMm}{r^2}$ $V_o = \sqrt{\frac{GM}{r^2}}$ $V_o = \sqrt{\frac{GM}{r}}$ If the test mass is at small distances from the source mass $r \approx R$ (radius of the source mass) Hence, $V_o = \sqrt{\frac{GM}{r}}$…………….(2) The above formula suggests that the orbital velocity is independent of the test mass (the mass which is orbiting). ### Relationship between Escape and Orbital Velocity The relationship between escape velocity and orbital velocity can be mathematically represented as: $V_e = \sqrt{2}V_o$ Meanwhile, if we divide Equation (1) and Equation (2), we get, $\frac{V_e}{V_o} = \frac{\sqrt{2GM/R}}{\sqrt{GM/R}} $ $\frac{V_e}{V_o} = \sqrt{2} $ It shows that escape velocity is $\sqrt{2}$ times greater than orbital velocity. ## The Motion of Satellites around Earth Let us consider a satellite assumed to be revolving around the earth in a circular orbit of radius 'r', having the centre of the earth as its centre; let us see the kinematics of the satellite moving around the earth. Here, the test mass is a satellite, and the source mass is earth. The velocity with which a satellite orbits around the earth is given by the orbital velocity, $V_o = \sqrt{\frac{GM}{r}}$ ### Time Period of a Satellite The time taken by the satellite to complete one revolution around the earth is known as the time period of a satellite. The time period of a satellite is the ratio of the total distance travelled by the satellite around the earth to the orbital velocity. $T = \frac{Total distance travelled by the satellite}{Orbital velocity}$ $T = \frac{2πr}{V_o}$ $T = \frac{2πr}{\sqrt{GM/r}}$ $T = \frac{2π}{\sqrt{GM}}(r)^{3/2}$ By squaring we get Kepler's third law $T^2 = \frac{4π²}{GM}(r)^{3}$ The constant of proportionality in the above equation depends only on the source mass but not on the test mass. ### Kinetic Energy of a Satellite Kinetic energy is the energy possessed by the body in motion (whether translational, rotational or a combination of both) in the case of the satellite orbiting around the earth. The satellite makes a pure rotational motion. In a pure rotation, the kinetic energy of the satellite is given by, $K = \frac{1}{2}mr^2ω^2$ Where, the angular velocity of a satellite is related to the time period of a satellite by a formula, $ω = \frac{2π}{T}$ Substituting $K = \frac{1}{2}mr^2(\frac{2π}{T})^2$ $K = \frac{1}{2}mr^2\frac{4π^2}{T^2}$ From Kepler’s third law, we know the time period of a satellite, and by substituting it in the above formula, we get $K = \frac{GMm}{2r}$ Kinetic energy can never be negative for any force. ### Potential Energy of a Satellite Potential energy is the energy possessed by the body in a particular position. Potential energy changes when the position of the body changes. To study potential energy, we require two bodies, one is source mass (*M*) which provides the gravitational force, and the other is test mass (*m*) which experiences the gravitational force by the source mass. In this case, the satellite is the test mass, and the earth is the source mass. The potential energy possessed by the satellite at a distance 'r' from the centre of the earth is given by, $U = \frac{-GMm}{r}$ ### Total Energy of a Satellite The total energy of the satellite is the sum of all energies possessed by the satellite in orbit around the earth. As only the mechanical motion of the satellite is considered, it has only kinetic and potential energies. Total energy of the satellite = kinetic energy of the satellite + potential energy of the satellite $ E = K + U$ $E = \frac{GMm}{2r} + \{\frac{-GMm}{r}\}$ $E = \frac{GMm}{2r}$ From the above equation, Total energy of the satellite = - (kinetic energy of the satellite) Total energy of the satellite = potential energy of the satellite ### Binding Energy of a Satellite Binding energy is the energy required to be given to the satellite to escape the gravitational pull of the earth. It is numerically equal to the negative of the total energy possessed by the satellite. Binding energy of a satellite = $\frac{GMm}{2r}$ The binding energy of a system of two bodies is the amount of minimum energy required to separate them by infinite distance, or it is simply the amount of energy required to make the potential energy of the system equal to zero, which is numerically equal to the kinetic energy of the body under motion. ### Angular Momentum of a Satellite When a satellite of mass 'm' is orbiting with an angular speed on the orbital path of radius 'r', its angular momentum is given by: $L = mr^2ω$ We know that $ω = \frac{2π}{T}$ substituting this in the above equation, we get $L = mr^2\frac{2π}{T}$ Using $T= \frac{2π}{\sqrt{GM}}(r)^{3/2}$, we get; $L = m\sqrt{GMr} $ From the above equation, it is clear that the angular momentum of a satellite depends on both the mass of a satellite and the mass of earth. It also depends upon the radius of the orbit of the satellite. ## Solved Examples on Chapter 6 1) Titan, the largest moon of Saturn, has a mean orbital radius of 1.22 × 10⁹m. The orbital period of Titan is 15.95 days. Hyperion, another moon of Saturn, orbits at a mean radius of 1.48 × 10⁹ m. Use Kepler's third law of planetary motion to predict the orbital period of Hyperion in days. **Solution** $r_T = 1.22 \times 10^9 m$ $T = 15.95 days$ $r_H = 1.48 \times 10^9 m$ $T_H = ?$ $ \{\frac{T_H}{T}\}^2 = \{\frac{r_H}{r_T}\}^3$ $ \{\frac{15.95 days}{T_H}\}^2 = \{\frac{1.22\times10^9m}{ 1.48\times10^9m}\}^3$ $ \{\frac{15.95 days}{T_H}\}^2 =0.824^3$ $ \frac{254.4 days^2}{T_H^2} = 0.560$ $T_H^2 = \frac{254.4 days^2}{0.560}$ $T_H = \sqrt{21.3 days^2}$ $T_H = 21.3 days$ 2) A bowling ball that has a weight of 2.2 kg falls 50 m to the ground below. Find the ball's gravitational potential energy when it reaches the bottom. **Solution** mass of the ball = 2.20 kg height of the ball = 50.00 m The gravitational potential energy is: $GPE = mgh$ $2.20 kg \times 9.80m/s \times 50.00m = 1078 J$ 3) If the mass of the earth is 5.98 × 10²⁴ kg and the mass of the sun is 1.99 × 10³⁰ kg and the earth is 160 million km away from the sun, calculate the GPE of earth. **Solution** Mass of the earth (*m*) = 5.98x10²⁴ Mass of the earth (*M*) = 1.99×10³⁰ The gravitational potential energy is given by: $U = \frac{GMm}{r}$ $U = \frac{(6.673×10^{-11}×5.98×10^{24}×1.99×10^{30})}{160×10^9}$ $U = 4963 \times 10^{30} J$ 4) Hydrogen escapes faster from the earth than oxygen. Why? **Solution** The thermal speed of hydrogen is much larger than oxygen. Therefore, a large number of hydrogen molecules can acquire escape velocity than that of oxygen molecules. Hence, hydrogen escapes faster from the earth than oxygen 5) A satellite is revolving round the earth in a circular orbit with critical velocity of 7 km/s. a) Find the radius of the orbit of the satellite and b) period of its resolution. (G = 6.673×10^-11 Nm2/kg2, M = 5.98x10²⁴ kg). **Solution** a) $v = 7 \times 10^3$ m/sec $V_o = \sqrt{\frac{GM}{r}}$ $7\times 10^3 = \sqrt{\frac{(6.673 \times 10^{-11} \times 5.98 \times 10^{24} \times 1.99 \times 10^{30})}{r}}$ $r = 8140122 m$ b) $T = \frac{2π}{\sqrt{GM}}(r)$ $T = \frac{2π}{\sqrt{GM}} \sqrt{r\times r}$ $T = \frac{2 π}{\sqrt{GM}} \sqrt{r\times r}$ $T = \frac{2π}{7 \times 10^3} \times r$ Therefore; $T = 7307 sec$