Gravitation & Kepler’s Laws What causes YOU to be pulled down to the surface of the earth? THE EARTH….or more specifically…the EARTH’S MASS. Anything that has MASS has a gravitational pull towards it.
푭품 M m What the proportionality above is saying is that for there to be a
FORCE DUE TO GRAVITY (Fg) on something there must be at least 2 masses involved, where one can be larger than the other.
The force of gravity (Fg) is directly proportional () to the masses (M and m) of the objects As you move AWAY from the earth, your DISTANCE increases and the FORCE DUE TO GRAVITY decreases. This is a special INVERSE relationship called an Inverse-Square. ퟏ 푭 품 풓ퟐ The “r” stands for SEPARATION DISTANCE and is the distance between the CENTERS OF MASS of the 2 objects. We use the symbol “r” as it often represents the radius. Gravitation is closely related to circular motion as you will discover later. Constant of Universal Masses of two objects Gravitation, being attracted to 6.67x1011 m³/kgs² each other m1m2 Fg = G Force of r2 Square of the distance attraction between the objects EXAMPLE: Two planets have the same masses but one is larger than the other. Neither planet has an atmosphere. Which of the following would be the same for objects of equal mass near the surfaces of the two planets? I. The final velocity of the objects if they are dropped from the same height. II. The momentum of the objects right before hitting the ground if they are dropped from the same height. III. The centripetal force acting on the masses if they are attached to identical strings and are spun in vertical circles with the same tangential velocities. A) I only B) II only C) I and II only D) II and III only E) None of the above are the same Since the force of gravity is different on each planet because the radii of the planets are different, the final velocities will be different v = 2gh, the momenta will be 풗ퟐ different p = mv, and the centripetal force will be different: F = m . c 풓 P.O.D. 1. Two planets have the same size but different masses and no atmospheres. Which of the following would be different for objects of equal mass on the surfaces of the two planets? I. The rate at which each would fall freely. II. The weight of the objects on weight scales. III. The distance that the objects stretch a spring vertically. A) I only B) II only C) I and II only D) II and III only E) I, II, and III
Gravity (g) vs. Force of attraction (Fg)? Let’s set the 2 equations for the force of attraction equal to each other since they BOTH represent your weight or force due to gravity
Fg mg Use this when you are on the earth m m F G 1 2 Use this when you are LEAVING the earth g r 2
Mm 11 24 (6.67x10 )(5.97x10 ) 2 mg G 2 g 9.81m / s r (6.37x106 )2 M g G SOLVED FOR g r 2 M Mass of the Earth 5.97x1024 kg, r dist. from center of the Earth 6.37x106 m Example How far from the earth's surface must an astronaut in space be if she is to feel a gravitational acceleration that is half what she would feel on the earth's surface? Mm mg G r 2 M M Mass of the Earth 5.97x1024 kg g G r 2 r radius of the Earth 6.37x106 m M 2 g G 2 g(r rearth ) G M (r rearth ) (6.67x1011)(5.97x1024 ) r 6.37x106 G M (r r )2 4.9 earth g
G M (r r )2 6 earth g r = 2.64 10 m G M r r earth g GM This value is four tenths the r Earth r g Earth radius of Earth. P.O.D. 2: A cat weighing 600 N on earth travels to a planet with 4 times the mass and twice the radius of Earth. The cat's weight on this planet is most nearly A) 300 N 800 B) N 2 C) 600 N D) 600 2 N E) 1200 N The Period of a Pendulum
Ans. The period of a pendulum is given by 푳 푻 = ퟐ흅 , 품 where T is the period (in s) for the pendulum to go back and forth, L is the length of the pendulum (in m), and g is the value of gravity (in m/s2) Example (a) What is the period of a pendulum on the earth’s surface if it has a length of 3 m?
Ans. The period of a pendulum 푳 is given by 푻 = ퟐ흅 품 S퐮퐛퐬퐭퐢퐭퐮퐭퐢퐧퐠 퐨퐮퐫 퐠퐢퐯퐞퐧 퐯퐚퐥퐮퐞퐬: 푳 ퟑ풎 푻 = ퟐ흅 = ퟐ흅 = ퟑ. ퟒퟖ 풔 품 ퟗ.ퟖ 풎 ൗ풔ퟐ Example (b) What is the period of the pendulum if it is at a height of 30 km?
First we need to find the value of g at 30 km = 30,000 m M M g G g (6.67x1011 ) r 2 r 2 M Mass of the Earth 5.97x1024 kg, r radius of the Earth 6.37x106 m
(6.671011 )(5.971024 ) g 9.72m / s2 (6.37106 30,000)2 S퐮퐛퐬퐭퐢퐭퐮퐭퐢퐧퐠 퐨퐮퐫 퐠퐢퐯퐞퐧 퐯퐚퐥퐮퐞퐬: 푳 ퟑ풎 푻 = ퟐ흅 = ퟐ흅 = ퟑ. ퟒퟗ 풔 Compared to 3.48 s on 품 ퟗ.ퟕퟐ 풎 ൗ풔ퟐ the previous slide P.O.D. 3. A simple pendulum has a period of approximately 2.0 s on earth. What is its period on an imaginary planet the same size as earth, but half the mass of earth? A) 0.5 s B) 1.0 s C) 1.4 s D) 2.0 s E) 2.8 s Energy Considerations Work is the integral of a Force function, F(r), with respect to displacement, r. REMEMBER: Work was the area under the curve of the Force vs. distance graph Substituting in the basic expression for gravitational force F = G푀∙푚 for F(r) g 푟2
Pulling out the constants and bringing the denominator to the numerator.
Integrating…
This formula gives you the work done by a mass against the variable force of gravity as the object gets further and further away from the planet … The negative sign should not be surprising as we already knew that Work was equal to the negative change in “U” or mgh. Example What is the radius of a planet if a rocket of mass 4000 kg requires 24,000 J of work to overcome the planet’s force of gravity? The planet has a mass of 3.6 × 1030 kg. mM mM W G r G r W M Mass of the Planet 3.6x1030 kg, m Mass of the rocket 4000 kg, G Universal Gravitation Constant 6.67x1011 N m2 kg2 W work done against gravity 24000 J (6.671011 Nm2kg2 )(4000kg)(3.61030 kg) r 4,002m 24,000J P.O.D. 4. A planet has a mass 4 times as great and a radius 2 times as great as that of Planet Earth. How many times more work would a rocket have to do to overcome the pull of gravity on it by the planet? A) The same B) Two times C) Four times D) Eight times E) Sixteen times Very far from center of earth: very Escape Velocity small vesc speed needed to escape an object’s gravitational pull.
Farther from center Speed greater than vesc. of earth: small vesc Consider a rocket leaving the earth. It Speeds smaller usually goes up, slows Surface of earth: than vesc. down, and then returns large v to earth. esc There exists an initial minimum speed that when reached the rockets will continue on forever. Let's use conservation of energy to analyze this situation!... Escape Speed
We know that ENERGY will never change. As the rocket leaves the earth it's kinetic is large and its potential is small. As it ascends, there is a transfer of energy such that the difference between the kinetic and potential will always equal to ZERO. Escape Speed
Set the potential energy equal to the kinetic energy and solve for v, to find the ESCAPE VELOCITY: P.O.D. 5. a. How many times greater is the Earth’s escape velocity compared to Mars’? b. How many times greater is Jupiter’s escape velocity compared to Earth’s? Newton’s Cannon Why don’t satellites fall to the earth if they are being pulled to it by gravity? •If the cannon shoots a cannon ball at a speed of say 8 m/s, the following will happen… It falls to earth pulled by gravity Newton’s Cannon Why don’t satellites fall to the earth if they are being pulled to it by gravity? •If the cannon shoots a cannon ball at say 80 m/s, the following will happen… Newton’s Cannon Why don’t satellites fall to the earth if they are being pulled to it by gravity? •If the cannon shoots a cannon ball at slightly less than 8000 m/s, the following will happen… The cannon ball will fall at the base of the mountain. Newton’s Cannon Why don’t satellites fall to the earth if they are being pulled to it by gravity? •If the cannon shoots a cannon ball at 8000 m/s, the following will happen… The cannon ball will orbit in a circular path Newton’s Cannon Why don’t satellites fall to the earth if they are being pulled to it by gravity? •If the cannon shoots a cannon ball at a velocity slightly over 8000 m/s, what will happen? The cannon ball will orbit in an elliptical path Why doesn’t the earth crash into the sun if it is attracted to it? Ans: the earth is always “falling” towards the sun. The reason it doesn’t crash into the earth is that it is moving fast enough that it always “misses”. The earth also has inertia that keeps it on its path. Kepler's Laws There are three laws that Johannes Kepler formulated when he was studying the heavens
THE LAW OF ORBITS - "All planets move in elliptical orbits, with the Sun at one focus.”
THE LAW OF AREAS - "A line that connects a planet to the sun sweeps out equal areas in the plane of the planet's orbit in equal times, that is, the rate dA/dt at which it sweeps out area A is constant.”
THE LAW OF PERIODS - "The square of the period of any planet is proportional to the cube of the semi major axis of its orbit." Kepler’s 1st law – The Law of Orbits "All planets move in elliptical orbits, with the Sun at one focus.” Kepler’s 2nd Law – The Law of Areas
"A line that connects a planet to the sun sweeps out equal areas in the plane of the planet's orbit in equal times, that is, the rate dA/dt at which it sweeps out area A is constant.”
A1 A2
A1 = A2 The planet moves faster near the Sun so the same area is swept out in a given time as at larger distances, where the planet moves more slowly. The green arrow represents the planet's velocity, and the purple arrows represents the force on the planet (the NET force, which is a combination of the centripetal force and the tangential force) Kepler’s 3rd Law – The Law of Periods "The square of the period of any planet is proportional to the cube of the semi major axis of its orbit."
Gravitational forces are centripetal, thus we can set them equal to each other!
Since we are moving in a circle we can substitute the appropriate velocity formula!
The expression in the RED circle derived by setting the centripetal force equal to the gravitational force is called ORBITAL SPEED.
Using algebra, you can see that everything in the parenthesis is CONSTANT. Thus the proportionality holds true! P. O. D. 6. A satellite is orbiting around the Earth at a radius of 2R from its center. What is the magnitude of the velocity it is traveling at? 퐺푀 A) 푅 퐺푀 B) 2푅 2퐺푅 C) 푀 D) 9.8 m/s 퐺 E) 푅 Kepler’s 3rd Law – The Law of Periods "The square of the period of any planet is proportional to the cube of the semi major axis of its orbit." ퟐ ퟑ 푻 풓 푨 = 푨 푻푩 풓푩
Where TB and rB are the period and semi major axis of the earth, which are 1 sidereal year and 1 풓 푨 astronomical unit, the standard. For planets with CIRCULAR orbits, the following version of the formula can also be used: ퟒ흅ퟐ 푻ퟐ = r3 푮푴 P. O. D. 7. Five planets make circular orbits around a star which is much more massive than any of them. The mass and orbital radius of each satellite is given. Which satellite has the longest period? A) Mass: 2M Radius: R B) Mass: M Radius: 2R C) Mass: M Radius: R D) Mass: 2M Radius: R E) Mass: M Radius: 2R P.O.D. 6 A satellite in orbit around the Earth has a period of one hour. An identical satellite is placed in an orbit having a radius which is nine times larger than the first satellite. What is the period of the second satellite? A) 0.004 hr B) 1/3 hr C) 3 hr D) 9 hr E) 27 hr Kinetic Energy in Orbit
Using our ORBITAL SPEED derived from K.T.L and the formula for kinetic energy we can define the kinetic energy of an object in a bit more detail when it is in orbit around a body.
The question is WHY? Why do we need a new equation for kinetic energy? Well, the answer is that greatly simplifies the math. If we use regular kinetic energy along with potential, we will need both the orbital velocity AND the orbital radius. In this case, we need only the orbital radius. Total Energy of an orbiting body
Notice the lack of velocities in this expression as mentioned in the last slide.
So by inspection we see that the kinetic energy function is always positive, the potential is negative and the total energy function is negative.
In fact the total energy equation is the negative inverse of the kinetic.
The negative is symbolic because it means that the mass “m” is BOUND to the mass of “M” and can never escape from it. It is called a BINDING ENERGY. Energy from a graphical perspective
As the radius of motion gets larger. The orbiting body’s kinetic energy must decrease ( slows down) and its potential energy must increase ( become less negative).
By saying become less negative means that we have defined our ZERO position for our potential energy at INFINITY. How do you move into a higher velocity orbit?
Question: If we have an orbiting Earth satellite and we want to put it in a higher velocity orbit, how can we use the satellite’s thrusters to make the adjustment?
a) Fire Backwards b) Fire Forwards
50% 50% Backwards = speed up Forwards = slow down
Fire Forwards Fire Backwards 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 How do you move into a higher velocity orbit?
1) If you fire backwards thinking you will speed up the satellite you put it into a larger orbital radius which ultimately SLOWS DOWN the satellite as the KE decreases. 2) By thrusting backwards you are ADDING energy to the system moving the total energy closer to ZERO, this results in a larger radius which also causes the KE to decrease. 3) Fire forwards gently so that you do NEGATIVE WORK. This will cause the satellite to fall into a smaller orbit increasing the KE and increasing the speed. It also makes the potential energy increase negatively because you are moving farther from infinity. As the potential increase the KE again decreases. 8. The Earth exerts the necessary centripetal force on an orbiting satellite to keep it moving in a circle at constant speed. Which of the following statements best explains why the speed of the satellite does not change even though there is a net force exerted on it? A) The satellite is in dynamic equilibrium. B) The acceleration on the satellite is zero. C) The centripetal force is in the direction of the velocity of the satellite. D) The centripetal force is equivalent to the reaction force caused from the satellite. E) The centripetal force is perpendicular to the velocity of the satellite. 10. A satellite orbits the Earth in at a distance of 1 Earth radius from the surface. Its velocity is v. If the satellite moves and enters a new orbit at 3 Earth radii from the Earth's surface, its new velocity will be 푣 A) 2 B) 푣 3 C)푣 2 푣 D) 3 푣 E) 2 FREE RESPONSE 1. In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data back to Earth. Assume a circular orbit with a period of 1.18 102 minutes = 7.08 103 s and orbital speed of 3.40 103 m/s. The mass of the GS is 930 kg and the radius of Mars is 3.43 106 m. (a) Calculate the radius of the GS orbit. (b) Calculate the mass of Mars. (c) Calculate the total mechanical energy of the GS in this orbit. (d) If the GS was to be placed in a lower circular orbit (closer to the surface of Mars), would the new orbital period of the GS be greater than or less than the given period? ______Greater than ______Less than Justify your answer. (e) In fact, the orbit the GS entered was slightly elliptical with its closest approach to Mars at 3.71 105 m above the surface and its furthest distance at 4.36 105 m above the surface. If the speed of the GS at closest approach is 3.40 103 m s , calculate the speed at the furthest point of the orbit. FREE RESPONSE 2. A student is given the set of orbital data for some of the moons of Saturn shown below and is asked to use the data to determine the mass MS of Saturn. Assume the orbits of these moons are circular. Orbital Period, T Orbital Radius, R (seconds) (meters) 8.14 104 1.85 108 1.18 105 2.38 108 1.63 105 2.95 108 2.37 105 3.77 108 (a) Write an algebraic expression for the gravitational force between Saturn and one of its moons. (b) Use your expression from part (a) and the assumption of circular orbits to derive an equation for the orbital period T of a moon as a function of its orbital radius R. (c) Which quantities should be graphed to yield a straight line whose slope could be used to determine Saturn’s mass? (d) Complete the data table by calculating the two quantities to be graphed. Label the top of each column, including units. (e) Plot the graph. Label the axes with the variables used and appropriate numbers to indicate the scale.