HW Solutions # 11-8.01 MIT - Pof. Kowalski Univesal Gavity. 1) 12.23 Escaping Fom Asteoid Please study example 12.5 "fom the eath to the moon". a) The escape velocity deived in the example (fom enegy consevation) is : 2GmA v esc = (1) R A Whee: m A Asteoid s mass=3.6 10 12 kg R A Asteoid s adius=700 m G =6.673 10 11 Nm 2 /kg 2 Plugging in these numbes into equation (1): v esc =0.83 m/s You can cetainly walk that fast on eath. Howeve, you could not walk on the asteoid faste and faste to achieve this speed because you would go into obit at a lowe velocity (v esc / 2) at which point the nomal foce of the gound would be zeo so thee would be no moe fiction to acceleate you! To get a feeling of how small this gavity is let s do some estimation of the time it takes to each this velocity on the asteoid. The gavitational foce F g = Gm A m/ra 2 on the suface of planet fo a mass, m 100 kg is 0.05 N. Let s take µ = 1. The fiction would be 0.05 N so the acceleation a 0.05/100=0.0005 m/s 2 and the time t it takes to each v esc =0.83 m/s is: t 0.83 = 1660 s 30 min 0.0005 1
b) The question is about the compaison with vetical leap on eath. Using: v 2 y v 2 0y = 2g eath d we have: d = v2 0y 2gd 0.03m d 3cm Even octogenaians can jump 10 this length. 2
2) 12.24 Satellite s altitude and mass m S := Satellite s mass. m E := Eath s mass. R := The distance between the cente of eath and satellite. F g :=the gavitational foce between the two masses. U := the gavitational enegy between the two masses. a) F g and U ae given. By witing them in tems of m S and m E : F g = Gm Em S R 2 (2) you can eliminate R: U = Gm Em S R (3) R = U F g (4) With the numbes given in the poblem (F=19.0 kn; U= 1.39 10 11 J) you ll get: R =7.31 10 6 m To find the altitude above the eath denoted by H you should subtact it fom Eath adius: H = R R E =7.31 10 6 6.38 10 6 =9.3 10 5 m H =9.3 10 5 m b) You can use the value of R and use eithe of (2) o (3) to find m S : Whee m E =5.97 10 24 kg: m S = RU Gm E m S =2.55 10 3 kg 3
3) 12.46 Gavitation fom 3 masses Let s use thee indices appopiate fo 3 masses namely: R ight the mass at (0.5m, 0): PR =0.5m; m R =1.0 kg. U p the mass at (0,0.5m): PU =0.5 m; m U =1.0 kg. D iagonal the mass at (0.5m,0.5 m): PD =0.5 2 m; m D =2.0 kg. a) Because of symmety we expect to get the total F acting on P along the diagonal. We it moe moe geneally though. The thee foces acting on mass P at oigin ae (F PR F acting on P fom R): F PD F PR =+ Gm P m R 2 R F PU =+ Gm P m U 2 U ˆx ŷ =+ Gm m P D ( ˆx + ŷ ) 2 D 2 Whee 2 comes fom the pojection of F D 45 with espect to x axis) on x and y axis. and F P (which is oiented at = F PR + F PU + F PD Using the values given in the poblem you ll get the magnitude: F P =9.67 10 12 N which oients along the diagonal with unit vecto ˆn 1 2 (ˆx + ŷ). F P =9.67 10 12 N ˆn 4
b) Use enegy consevation K 2 + U 2 = K 1 + U 1 (5) whee 1 denotes the situation that mass P is 300m fa away fom oigin. U 1 is pactically zeo at this lage distance. (this distance is 2 odes of magnitude lage so within 1% appoximation you can ignoe U 1 : K 1 =0 U 1 0 U 2 = Gm P ( m R PR + m U PU U 2 + 1 2 m P v2 2 0 + m D PD ) The facto of m P v 2 = will be cancelled: v 2 = 2G( m R PR 2U 2 m P + m U PU + m D PD ) (6) Plugging the given numbes into (6): v 2 =30.2 ± 0.3 µm/s v 2 30 µm/s NOTE: Gavity is a vey weak foce (e.g. compaed with electostatic foces), even if it had the speed v 2 fo the entie jouney, it would take 1 yea to make this tip. In In fact it will take 1 yea if no othe foces (e.g. fom sunlight foces) come into play. 5
3) 12.68 Gavitational Potential a) Fom the definition given the units of φ is the unit of enegy divided by mass.we use the convention [ ] to denote the units: [φ] = [U] [m] = [m][v2 ] [m] =[v 2 ]=m 2 /s 2 b) The gavitational potential enegy of two masses m and m E sepaated by a distance (assuming Zeo enegy at infinity sepaation) is: Fom the definition we get: U = Gm Em φ = U m (7) φ() = Gm E c) The poblem asks fo the quantity φ = φ(r E + H) φ(r E ). Whee H denotes the altitude of the space station (400 km in this poblem). Using: G =6.673 10 11 Nm 2 /kg 2. m E =5.97 10 24 kg. R E =6.38 10 6 m. H = 400 10 3 m. You ll get: φ =3.68 10 6 m 2 /s 2 6
d) If you assume that the initial and final velocities ae Zeo which is equivalent to a vey gadual pocess. Using K i + U i + W lift = K f + U f Hee K i = K f =0andW lift is the wok that must be done against the gavity: W lift = U f U i = m φ Using m=15,000 kg and the esult of pat (c) you ll get: U =5.53 10 10 J e) To dock, you have to get up to the speed of the obiting space station. So the K f should be its final Kinetic enegy as an obiting payload. Fo a cicula obit we have: K f = U * f 2. Now going though the same pocedue as pat d but with K f = U f : 2 W obit = U f U i + K f = U f U i U f 2 =+U f 2 U i W obit W lift = U f U i + K f U f U i U i U f = f i = + Uf 2 U i U f U i = 1 2 U i U f 1 U i U f = R E + H R E =1+ H R E =1.06 W obit W lift = 0.5 1.06 1 1.06 =9.33 So getting thee is only 11% of the wok- most is getting up to obit speed. *: F = ma Gm Em 2 = mv2 K = 1 2 mv2 = Gm Em 2 = U 2 7
3) 12.70 Effect of Ai Dag on Satellite s Motion a) In moving to a lowe obit by whateve means, gavity does positive wok, and so the speed does incease. b) Fom Calculus you know that fo a geneal function f() we have: df () f( ) =f() d () f = df (8) d Whee is much smalle than ( 1). Fom the expession fo v() incicula motion: GmE v() = d( α ) = α α 1 (9) d Combined with (8) you ll get: GmE v =+( /2) > 0 3 Combining K =1/2mv 2 with (8) and (9) you ll get: K =+ Gm Em >0 2 2 Combining U == Gm Em with (8) and (9) you ll get: U = Gm Em 2 8
Total enegy is E = K + U so: U = Gm Em = 2 K 2 E = K + U = K 2 K = K W = E = K <0 c) Since =50 km is much smalle than R E itself so you can safely use equation (8) fo functions v, K and U. Fo the est you should just plug in the numbes and use: = R E + H =6.38 10 6 + 300 10 3 =6.68 10 6 m. =50 10 3 m. m = 3000 kg. v = GmE =7.72 10 3 m/s v =+( /2) GmE 3 E = K + U = Gm Em 2 = +28.9 m/s = 8.95 10 10 J K =+ Gm Em =6.70 10 8 J 2 2 U = 2 K = 1.34 10 9 J W = K = 6.70 10 8 J As the tem buns up suggests, the enegy is conveted to heat o is dissipated in the collisions of the debis with the gound. 9