Gravitational Potential energy Mr. Burns SPH4UI Gravitational Potential energy Mr. Burns

Gravitational Potential Energy To explore how much energy a spacecraft needs to escape from Earth’s gravity, we must expand on the topic of gravitational potential energy. To calculate the change in gravitational potential energy that a mass undergoes when it undergoes a vertical displacement near the Earth’s surface, we used: Where is the change in gravitational potential energy, m is the mass, g is the magnitude of the gravitational field constant, and is the vertical displacement, where when we choose hi to be the ground, we simply obtain U=mgh. This equation assumes that g remains reasonably constant during the change in vertical displacement. We define the work done on a system by a specific conservative force as the negative of the change in a potential energy function associated with the force. Changes are always taken as final minus initial values

Gravitational Potential Energy As we recall from the Kinetic energy Theorem: A force does positive work when it has a vector component in the same direction as the displacement. If the work done on a particle is positive, then the particles kinetic energy increases. Since the work done a we move a mass between two points is independent upon the path, the force is conservative

Gravitational Potential Energy Change in potential energy is equal to the negative of the work done by the gravitational force on a mass Calculus definition of work This gives us a Calculus definition of change in potential energy Equation for change in potential energy If we set, hi to be zero (the ground), we obtain an simple formula to determine the potential energy

Gravitational Potential Energy Work done by the gravitational force FG during the vertical motion of a mass from initial height y1 to final height y2 as the body rises. The gravitational potential energy U increases (negative work done by gravity, Kinetic Energy decreases). Work done by the gravitational force FG during the vertical motion of a mass from initial height y1 to final height y2 as the body rises. The gravitational potential energy U decreases (positive work done by gravity, Kinetic Energy increases)

Gravitational Potential Energy But, how do we handle Gravitational Potential Energy, when g is not a constant (such as when we are not on Earth)?

Gravitational Potential Energy Recall that the law of universal gravitation is given by: Where is the magnitude of the force of gravitational attraction between any two objects, M is the mass of one object, m is the mass of the other object, and r is the distance between the centres of the two To increase the separation of the two masses requires work to be done. That is, we must do work to overcome the gravitational attraction between the two masses when moving mass m from distance r1 to distance r2 r1 M m r2

Gravitational Potential Energy As the result of the work being done to increase the separation from r1 to r2, the gravitational potential energy of the system increases. The work done to change the separation from r1 to r2 is equal is equal to the change in gravitational energy from r1 to r2. However, recall that work done by a varying force is equal to the area under the force-displacement graph for that interval.

Gravitational Potential Energy To find an expression for this work, we consider a body of mass m outside the Earth and compare the work Wgrav done by the gravitational force when a body moves directly away from or towards the centre of the Earth (from r=r1 to r=r2). Because the Force points directly toward the centre of the Earth, F is negative The path does not have to be a straight line, it could also be a curve, but the work done only depends upon the initial and final values of r, not the path taken (this is definition of a conservative force). We now define the corresponding potential energy U so that Wgrav=U1-U2. Therefore we have a final definition of gravitational potential energy

Gravitational Potential Energy You may be troubled by the equation for gravitational potential energy because it states that it is always negative. But we have encountered this before, recall when we used U=mgh, we found that U could be negative whenever the body of mass was below the arbitrary height we chose h to be zero at. In using our equation, we have chosen U to be zero when the body of mass is infinitely far away from the Earth . As the body moves towards the Earth, gravitational potential energy decreases and becomes negative.

Gravitational Potential Energy These formula seen initially confusing when first encountered. When gravity field is constant (near Earth) Universal Potential energy, Equals zero when r approaches infinity But, they provide us with the same meaning. The change in gravitational energy at the Earth's Surface is just a special case of the general solution. This is the gravity well scientists talk about.

Note: h is distance above the ground Gravitational Potential Energy What is the change in gravitational potential energy of a 72.1 kg astronaut, lifted from Earth’s surface into a circular orbit at an altitude of 4.35 x 102 km? Note: h is distance above the ground Let’s approximate the answer using U=mgh

This is the escape velocity of the mass Gravitational Potential Energy We can use the fact that if no other external forces besides gravity affect a mass, then the total energy is conserved (the gravitational force does work and thus the mechanical energy is conserved). The Kinetic Energy gained by the system is matched with the Potential Energy lost, and thus our mass will have the following statement of energy conservation.. Now by inserting the formulas for the Kinetic and Potential Energy and interpreting the scenario where the object reaches the r location where both EK2 and U2 have a zero value (where the object comes to a stop). This gives us the initial speed v1, needed for a body to escape from the surface of a spherical mass M with radius r This is the escape velocity of the mass

Gravitational Potential Energy Suppose you want to place a 1000 kg weather satellite into a circular orbit 300 km above Earth’s surface. a) What speed, period, and radial acceleration must it have? b) How much work has to be done to place the satellite in orbit? c) How much additional work must be done to make this satellite escape Earth’s gravity ME=5.97x1024 kg RE=6380 km First we need the radius of the satellites orbit:

Gravitational Potential Energy Suppose you want to place a 1000 kg weather satellite into a circular orbit 300 km above Earth’s surface. a) What speed, period, and radial acceleration must it have? b) How much work has to be done to place the satellite in orbit? c) How much additional work must be done to make this satellite escape Earth’s gravity For a stable orbit: FG=mac

Gravitational Potential Energy Suppose you want to place a 1000 kg weather satellite into a circular orbit 300 km above Earth’s surface. a) What speed, period, and radial acceleration must it have? b) How much work has to be done to place the satellite in orbit? c) How much additional work must be done to make this satellite escape Earth’s gravity For a stable orbit: FG=mac

Gravitational Potential Energy Suppose you want to place a 1000 kg weather satellite into a circular orbit 300 km above Earth’s surface. a) What speed, period, and radial acceleration must it have? b) How much work has to be done to place the satellite in orbit? c) How much additional work must be done to make this satellite escape Earth’s gravity

Gravitational Potential Energy b) How much work has to be done to place the satellite in orbit? The work required is the difference between the total mechanical energy when the satellite is in orbit (Ef) and when the satellite was on the launch pad (Ei). RE=6380 km R =6680 km ME=5.97x1024 kg

Gravitational Potential Energy c) How much additional work must be done to make this satellite escape Earth’s gravity We recall that for a satellite to escape to infinity, the total mechanical energy must be zero. The total mechanical energy in orbit Ef was -2.99 x 1010 J To increase this to zero, an amount of work equal to 2.99 x 1010 J must be done.

Gravitational Potential Energy A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous orbit (one orbit is 24 hours) around Earth .   a) What is the radius of the satellite’s orbit? b) What is the velocity of the satellites orbit? c) What is the gravitational potential energy (universal) of the satellite when it is at rest on Earth’s surface? d) What is the total energy of the satellite when it is in geosynchronous orbit? e) How much work the launch rocket do on the satellite to place it into orbit? f) Once in orbit, how much additional energy would the satellite require to escape from Earth’s potential well? g) What should the launch velocity be it the satellite is required to escape from Earth’s potential well?

Gravitational Potential Energy A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous orbit (one orbit is 24 hours) around Earth .   What is the radius of the satellite’s orbit? Since we are given the period of the orbit is 24 hours. We will use the Period version of centripetal acceleration.

Gravitational Potential Energy A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous orbit (one orbit is 24 hours) around Earth .   b) What is the velocity of the satellites orbit?

Gravitational Potential Energy A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous orbit (one orbit is 24 hours) around Earth .   c) What is the gravitational potential energy (universal) of the satellite when it is at rest on Earth’s surface?

Gravitational Potential Energy A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous orbit (one orbit is 24 hours) around Earth .   d) What is the total energy of the satellite when it is in geosynchronous orbit?

Gravitational Potential Energy A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous orbit (one orbit is 24 hours) around Earth .   e) How much work does the launch rocket do on the satellite to place it into orbit?

Gravitational Potential Energy A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous orbit (one orbit is 24 hours) around Earth .   f) Once in orbit, how much additional energy would the satellite require to escape from Earth’s potential well? The negative of Total Energy to get the satellite there.

Gravitational Potential Energy A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous orbit (one orbit is 24 hours) around Earth .   g) What should the launch velocity be if the satellite is required to escape from Earth’s potential well?

Understanding Work & Energy A rock is dropped from a distance RE above the surface of the earth, and is observed to have kinetic energy K1 when it hits the ground. An identical rock is dropped from twice the height (2RE) above the earth’s surface and has kinetic energy K2 when it hits. RE is the radius of the earth. What is K2 / K1? The easiest way to solve this problem is to use the W=K property. (a) (b) (c) 2 2RE Be careful! RE RE

Since energy is conserved, Solution Since energy is conserved, Do not use mgh formula as this only works when h is very small. For the first rock: For the second rock: 2RE RE RE

Kepler The planet Earth is at an average distance of 1 AU from the Sun and has an orbital period of 1 year. Jupiter orbits the Sun at approximately 5 AU’s. How long is the orbital period of Jupiter? Use Kepler’s Law T2=a3

Orbit Velocity Pluto’s orbital distance is 5.9x109 km. Given that the mass of the Sun is 2.0x1030 kg, and the mass of Pluto is 1.3x1022 kg. Determine the gravitational force the Sun exerts on Pluto. Use:

Multi-multiple Choice The Planet Mars orbits the sun at a nearly constant speed. Which of the following Statements are true?. There is a force on Mars towards the centre of the orbit. There is a force on Mars pulling it out from the centre of orbit There is a force on Mars in the direction of its motion There is acceleration on Mars toward the centre of orbit There are no forces on Mars.

g The Moon takes 27.3 days to orbit the Earth at an average distance of 385,000 km from the centre of the Earth. What is the acceleration of the Moon? 1.54 x 10-3 m/s2 2.73 x 10-3 m/s2 5.84 x 10-3 m/s2 7.51 x 10-3 m/s2 9.82 x 10-3 m/s2 Use Centripetal Force = Gravitational force to solve for ‘a’

g If in the previous question we assign the acceleration of the Moon with am and the mass of the Moon with Mm. If we place an astronaut with a mass Ma in the same orbit with the same speed, what would be gravitational field strength would the astronaut feel (Ma/Mm)am (Mm/Ma)am (Mm+Ma)am (Mm-Ma)am am Use Centripetal Force = Gravitational force to solve for ‘a’