Given the following equation for orbital velocity, pe sin 0f + (1 + e cos 0) Ô demonstrate mathematically that the magnitude of orbital velocity reaches its maximum at periapsis.
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- I am questioning the r value here: from where did you arrive to (2.23×10)24. Also, am I pluging the equation cortrectly here?? Can youy help, please - thanks T2 = {[(4) (π)2] [(G) (M)]} (r)3 M = {[(4) (π)2] / (G) (T)2]} (r)3 M = (4) (3.14)2 / (2.23×10)24 (6.67×10)-11 (57,888)2 M = 1.88×1027 kgwhat is the answer for this qustion (a) ?where does 3.6 come from?
- B4In spherical coordinates, the ladder operators for orbital angular momentum are of the form: i. - * * L4 eig +i cot 0- +i cot 0- Show, by explicit calculation of the relevant products, that these operators satisfy the commutation relations [L°, L4] = 0. [L4, L_] = 2L,.Our solar system orbits the center of the Milky Way galaxy. Assuming a circular orbit 30,000 ly in radius and an orbital speed of 250 km/s, how many years does it take for one revolution? Note that this is approximate, assuming constant speed and circular orbit, but it is representative of the time for our system and local stars to make one revolution around the galaxy.
- The qs was solved here.urgent plsx(t) = a cos(wt) + b exp(-t) is a recurrent orbit O True O FalseThe churge deisity asphre of glvn काि hब ऊगर पु pc,@,¢)= Krcos (20) sind pus (oe (Kunstant) of the sphee कdड मिण्र नजय cos 2XSinxdx= Te(-4x+4Sin&x-5h 4x calulute the t p ot dipok manut may need the followilag intyoad you nou b. Find the leading exparsion af the potarial at points r away from term in the multipok the