Variation of g with Height An object of mass m is on the surface of the Earth (mass M) where the weight of the object can be expressed with mg = GMm /r2 g = GM / r2… thus g a 1/ r2 above the Earth’s surface The density of the Earth (r = m/v) is not uniform and so this causes an unusual variation of g with radii inside the Earth 10 May 2018 Mechanics (Gravitation) Lesson #14

Gravitational Potential When a mass is lifted up in a gravitational field (moved against the field direction) work has to be done by an external agency and the mass gains gravitational potential energy. If it is released the field does the work and the potential energy is converted into kinetic energy. In order to measure potential energy at a point we need to find a zero point. We first define gravitational potential. 10 May 2018 Mechanics (Gravitation) Lesson #14

Gravitational Potential The gravitational potential at a point (V) in a gravitational field is defined as the work done by external forces in moving unit mass (m) from infinity to that point V = work done / mass … units of Jkg-1 The theoretical position of zero gravitational potential is taken to be at infinity. 10 May 2018 Mechanics (Gravitation) Lesson #14

Gravitational Potential The gravitational potential at a distance r from a point mass m is given by: V = - Gm r 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Example Calculate the gravitational potential due to the Sun at the orbit of Pluto (5.92x1012m). Mass V = - Gm r = - 6.67x10-11 x 1.99x1030 5.92x1012 = - 2.24x107 Jkg-1 10 May 2018 Mechanics (Gravitation) Lesson #14

The Potential ‘Well’ of the Earth This graph gives an indication of how masses are ‘trapped’ in the Earth’s field -20 2 4 6 8 -40 -60 Distance from Earth (in Earth radii) Vp (in MJ kg-1 ) The closer one gets to the mass m, the more negative the potential becomes. 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Conservative Field Gravity is known as a conservative force field as the work done by the force on the particle moving through a round trip is zero (energy is conserved) Ball thrown upwards with Ek returns to the senders hand with same Ek if air resistance is negligible By this argument a non-conservative force field is one which causes the energy of the system to change Friction and air resistance are labelled non-conservative 10 May 2018 Mechanics (Gravitation) Lesson #14

Gravitational Potential Energy Since the gravitational potential is defined as the work done moving unit mass from infinity to a point, the potential energy of a mass m at this point is given by: Ep = V x m Ep = - GMm r Note: As a consequence of defining the zero of gravitational potential to be at infinity. All the gravitational potential energy must be negative since work has to be done to get a mass to the point of zero energy. 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Changes in Potential Energy For large distances from the Earth, the change in potential energy can only be calculated using -GMm/R - -GMm/R’ where R’ and R are the distances from the centre of the Earth. For small distances above the Earth, the gravitational force on a mass is fairly constant. The change in potential energy can thus be calculated using mgh - mgh’ 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Example A spaceship of mass 4x105kg is at a distance 3x106m from the moon’s centre, when it fires it’s engines and moves out to a distance of 3.4x106m. Calculate the work done by the engines. Ew = Ep (gained) = final Ep – initial Ep = - GMm – - GMm r2 r1 = GMm ( 1/ r1 – 1/r2) = 6.67x10-11 x 7.3x1022 x 4x105 (1/3x106 – 1/3.4x106) = 7.64x1010J 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Satellite Motion Satellite motion can be considered as an extension of projectile motion. The combination of a horizontal motion, constant speed, and a vertical motion, constant acceleration, allows the satellite to complete a circular orbit. The fact that a satellite completes a circular path means that it must experience a centripetal force. Centripetal Force = Gravitational Force mv2/r = GMm/r2 v2 = GM/r v = (GM/r) The speed of a satellite is therefore determined by the radius of its orbit and is unaffected by the mass of the satellite. 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Period of a Satellite For a satellite in a circular orbit Centripetal Force = Gravitational Force mrω2 = GMm/r2 ω2 = GM/r3 but ω = 2π/T 4π2/T2 = GM/r3 T2 = 4π2r3/GM T = √(4π2r3/GM) T = 2π√(r3/GM) Note that the period of the satellite is independent of its mass. 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Energy of a Satellite A satellite of mass m in orbit round the Earth has both kinetic energy and potential energy. Ek = ½ mv2 where v is the speed in the orbit If the orbit of the satellite is circular, then if M is the mass of the Earth: mv2/r = GMm/r2 mv2 = GMm/r Hence Ek = GMm/2r Ep = -GMm/r i.e. the potential energy of the mass is numerically twice its kinetic energy. 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Therefore the total energy associated with the satellite is ET = Ep + Ek = -GMm/r + GMm/2r ET = -GMm/2r 10 May 2018 Mechanics (Gravitation) Lesson #14

Escape Velocity Derivation The Escape Velocity (ve) of a mass (m) on the surface of a planet of mass (M) is the minimum velocity needed for the mass to escape completely from the gravitational field of the planet i.e. to escape to infinity. Ek = Ep ½ mv2 - 0 = 0 - (-GMm/r) ½ mv2 = GMm/r v2 = 2GM/r v = (2GM/r) Note: The Escape Velocity does not depend on the mass of the object. 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Black Holes A black hole is formed when a sufficiently large star has reached the end of its life and explodes. The remains of the star then collapse inwards and the resultant body has an extremely large density. This dense body gives rise to a gravitational field which is so large that the escape velocity required to remove an object from that field would exceed the speed of light. According to Einstein’s theories, it is impossible for any body to exceed the speed of light and therefore no light can be observed exiting such a body. Hence the term ‘black hole’. http://hubblesite.org/explore_astronomy/black_holes/index.html 10 May 2018 Mechanics (Gravitation) Lesson #14

Black Holes and Photons Anything, including light, which gets too close to a black hole will be ‘sucked in’ by the gravitational field and will be unable to escape. Since photons passing a star are deflected by the star, stellar objects behind the star appear at slightly different positions as a result of the photons bent path. 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Satellite Problem A communications satellite of mass 4300 kg moves at constant speed at 2300 km above Earths surface Calculate (a) period of orbit (b) total energy at this height Mass Earth = 6x1024 kg G = 6.67x10-11 Nm2kg-2 Radius of Earth = 6.4x106 m 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Satellite Solution (a) T = 2p √( r3 / GM ) T = 8059.7 seconds (b) Etot = -GMm / 2r Etot = - 9.89x1010 Joules 10 May 2018 Mechanics (Gravitation) Lesson #14

Escape Velocity Problem Using standard data calculate the escape velocity of the Earth 10 May 2018 Mechanics (Gravitation) Lesson #14

Escape Velocity Solution ve = √( 2GM / r ) ve = 11183.13 ms-1 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Rocket Problem A rocket of mass 500 kg is fired into space from Earth and reaches a vertical altitude of 87000 km Calculate (a) Potential energy at the highest point (b) Calculate the return velocity of this rocket as it crashes into Earths atmosphere (which is approx 125 km up) 10 May 2018 Mechanics (Gravitation) Lesson #14

Mechanics (Gravitation) Lesson #14 Rocket Solution At 87000 km altitude Ep = -GMm / r = - 2.1424x109 J At 125 km altitude Ep = -GMm / r = - 3.067x1010 J Kinetic energy = DEp = ½ mv2 v = 10682.2 ms-1 10 May 2018 Mechanics (Gravitation) Lesson #14