1 PHY2048 Mach 31, 2006 Escape Velocity Newton s law of gavity: F G = Gm 1m 2 2, whee G = 667 10 11 N m 2 /kg 2 2 3 10 10 N m 2 /kg 2 is Newton s Gavitational Constant Useful facts: R E = 6 10 6 m M E = 6 10 24 kg Fo an obit at the Eath s suface v obit = GM E /R E 8 km/s Escape velocity off the suface of the Eath v esc = E /R E 12 km/s Obit of Pluto 5 1 2 lt-hs Example: 5kg mass is 1m away fom a 1kg mass and a 4kg mass The net gavitational foce on fom the 5kg mass fom the othe masses is zeo What is the distance between the 5kg mass and the 4kg mass? ns: 2m Gavitational Potential Enegy: The change in potential enegy is defined by Fo Newton s gavity, this gives U B U = U B = U B U = B GMm 2 d = B [ GMm ] B F dx = GMm B + GMm Potential enegy is only defined up to the addition of a constant It is natual (to physicists) fo the gavitational potential enegy outside an object to be U = GMm (1) Potential enegy is zeo when the objects ae infinitely fa apat
(2) Gavitational potential enegy deceases, and is negative, when the objects move close togethe 2 Example: If a ock is dopped fom a height of R E above the suface of the Eath, then how fast is it moving when it hits the Eath? ns: Fom the consevation of enegy K + U = 0 GMm 2R E = 1 2 mv2 GMm R E So v 2 = GM R E and v at the suface of the Eath is about 8 km/s cicula obit Fom F = m a we have fo a satellite, of mass m, in a cicula obit of adius aound an object of mass M ma = m v2 = F = GMm, 2 nd the speed is vob 2 = GM GM o v ob = which is called the obital speed v ob fo the Eath is about 8km/s Fo compaison, the otation of the Eath gives a point on equato a speed of about 04 km/s The kinetic enegy of the satellite is then kinetic enegy = K = 1 2 mv2 = GMm, 2 and the potential enegy is (fom two equations up) potential enegy = U = GMm Theefoe, the total enegy of a satellite in a cicula obit is tot enegy = K + U = 1 2 mv2 ob GMm ob = GMm 2 ob GMm ob = GMm 2 ob
Blast a satellite staight up: If a satellite, of mass m, goes staight up with a speed v 0 fom the suface of a planet of adius and mass M, then its speed will decease as gavity pulls back on the satellite But, the total enegy of the satellite is total enegy = K + U = 1 2 mv2 0 GMm, and this total enegy does not change So afte the satellite has moved to a geate distance away fom the planet, its total enegy is K + U = 1 2 mv2 GMm = 1 2 mv2 0 GMm We can use this to find the speed as a function of the distance, its initial distance, and its initial speed v 0, v 2 = v 2 0 3 The satellite eaches its futhest distance max when v = 0, so 0 = v 2 0 max O, we can find max as a function of the satellite s initial distance, and its initial speed v 0, max = 1 / v0 2 But what if v0 2 = /? Then, max =, and the satellite leaves the planet entiely Notice that as the satellite moves futhe away it is always slowing down Its speed appoaches zeo, but it neve quite stops The speed needed to escape completely is the escape velocity v esc Fo a stating distance, this is v esc = It is convenient to notice that v esc = 2v ob Fo the Eath, v esc is about 11 km/s
If a satellite stats out with a speed geate than the escape velocity then it neve eaches a place whee its speed is zeo and it stats to fall back down nd, in fact, as the satellite moves futhe and futhe away fom the planet, its speed does not decease to zeo bove, we used the consevation of enegy to show that So if v 0 > v esc = v 2 = v 2 0 then as v v 2 0 In this case, not only does the satellite get infinitely fa fom the planet, but it still has a finite speed when it gets thee 4 Example: satellite blasts off fom the suface of the Eath with an initial speed v 0 = 2v esc How fast is the satellite moving when it gets vey fa fom the Eath? ns: v = v 2 0 fo any distance When is vey lage and v 0 = 2v esc v = v0 2 = 4vesc R 2 = 4vesc 0 R 2 vesc 2 = 3vesc, 2 0 Finally, v = 3v esc, Black Holes Einstein s Special Theoy of Relativity implies that no object can tavel faste than the speed of light, c If thee wee an object whose escape velocity wee geate than the speed of light, then light could not escape fom it and we couldn t see it, so it would appea black lso, nothing can go faste than light, so nothing else could escape fom it eithe Such an object would be like a hole Such an object is called a black hole
What is the size of a black hole? Fom the equation above fo the escape velocity, we can set v esc = c and solve the equation fo the adius v 2 esc = c 2 = so that R S = c 2 R S is called the Schwazschild adius, afte a well-known Geman astonome fom the ealy 1900 s This is the adius of a nonotating black hole whose mass is M 5 black hole the mass of the sun has R S = a few kilometes black hole the mass of the Eath has R S = a few centimetes black hole the mass of a mountain has R S = about the size of a poton black hole the mass of the Univese has R S = about the size of the Univese It is a consevative point of view in astonomy today to believe that thee ae a numbe of black holes in the univese which is compaable to the numbe of stas these have a mass which is a few time the mass of ou sun lso many galaxies have jumbo sized black holes at thei centes with a mass anywhee fom 1,000 times the mass of the sun to 10 10 the mass of the sun Thee appeas to be a black hole in the cente of ou own Milky Way Galaxy which has a mass about 1,000,000 times the mass of the sun