Question

PART 1: Asteroids X, Y, and Z have equal mass of 7.0 kg each. They orbit around a planet with M=7.20E+24 kg. The orbits are in the plane of the paper and are drawn to scale.

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In the statements below, TE is the total mechanical energy, KE is the kinetic energy, and PE is the potential energy.  

greater than less than equal to  The PE of Z at i is .... the PE of Y at r
greater than less than equal to  The TE of Y is .... the TE of Z
greater than less than equal to  The PE of Z at i is .... the PE of X at a
greater than less than equal to  The KE of Y at i is .... that at r
greater than less than equal to  The TE of Z is .... the TE of X
greater than less than equal to  The speed of X at a is .... that at s
greater than less than equal to  The PE of Z at a is .... that at m
greater than less than equal to  The PE of Z at a is .... the PE of X at a

PART 2:

Asteroids X, Y, and Z have equal mass of 8.0 kg each. They orbit around a planet with M = 3.20×10²⁴ kg. The orbits are in the plane of the paper and are drawn to scale.
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The three asteroids orbit in the same clockwise direction.

greater than less than equal to  The period of X is .... that of Y.
greater than less than equal to  The angular momentum of Z at s is .... that at i.
greater than less than equal to  The angular momentum of Y is .... that of Z.
greater than less than equal to  The period of Z is .... that of Y.
greater than less than equal to  The angular velocity of Y at n is .... that at i.
greater than less than equal to  The angular velocity of Z at c is .... that at i.
greater than less than equal to  The angular velocity of Z at i is .... that of Y at i.

Answer 1

part 1

is. mass of asteroid each m = 7 kg. mass of the planet is M = 6.20 x 20 94 kg. spuud of the asturoud in ourbit is Gm = the of of asteroid is energy energy now kinetic KE - 1 mo? = 1 m (samo) GM m KE = GMM - 1 KE 201 Newton's law of and from guiquitation -GM m PE = u

now total energy is TE + РЕ KE = - Gm GMm TE = TE - २ज TE = – GMM (1) 2u7 forom equation (is , (i) & (iii) KE & I because of negative } PE & ut TE & ut 819 m sign ( (9) Four zo orbit Z - Four Ourbit sic > (KE) 2 c < (KE)z, u less than

(6) From the figure sly > ux (TE) > (TE), (TEx < (TE)y less than I (c) Four y ourbital utu < Ji greater than 1d) Four z and X oubital x > utz (TE), < (TEX I Less than Less than

(e) four x - oubital slu > uts (PE) 4 → (PE) x, s quuatur than I t For Y and a curbitals su <ule (PE), < (PE)z, less tham

119 I Four X and Y oubitals sturx - Story (PE) x 4 = (PE) y u X, 4 (h) Du л элс (PE)yy > (PE)x, I guenter tham I qruatur than
part2

Mass of each asturoid is m = 8 kg. mass of the planet is M = 5.20 X 1024 kg: oubital law of quinvitation from Newton's velocity is from the planet. fourmela where it be the distance now time period is from the distance = speed a time RITUT = vxT 2T UT T =

2T RIT 211 7 3/2 T T = GM JGM Tau angular velocity - 1 wa en from the figurer we can see that > uz > Jy now ourder of angular velocity is وانا < لیا بلا

four z oubital In > ste aut n is (a) The angular The angular velocity of z less than that at i . é (6) For 2 , aut point same so FOUT 2 and 7 oubital distance from the planet is Wze = wye equal to four y (c) curbital = wys = Wye equal to

(d) utz > Ty Tz > Ty I a greater than is Angular momentum ฤm งง L = L = m GMT = m GMG Lau . الحال Lz > Ly gerater than quruatur tham

momentum Since 97 (f) since anaular momentum of the planet always conserv so LzV = Lze equal to (1) I , 2 ly Tx > Ty Ty < Tx than