Question

A Planet revolves around the sun in an ellipse orbit of semi major axis 2x$10^{12}$m.The areal velocity of the planet when it is nearest to the sun is 4.4x$10^{10}$m/s.The least distance between planet and the sun is 1.8x10$10^{12}$m.Find the minimum speed of the planet in Km/s?.

Answer 1

See diagram:

AP = major axis of the elliptical orbit of the planet = 2 * 2 * 10¹² m

Nearest distance at Apihelon = AF = 1.8 * 10¹² m = Ra

The farthest distance at perihelon = PF = AP - AF = (4 - 1.8) 10¹² m
              PF = 2.2 * 10¹² m = Rp

Let us denote the linear (aerial) velocity of the planet as Vp at Perihelon P and as Va  at Apihelon A.

The angular momentum of the planet around the Sun is conserved. Hence
 
        M Rp Vp = M Ra Va
  
         Vp = Ra Va / Rp = 1.8 * 10¹² * 4.4 * 10¹° / 2.2 * 10¹² 
 
  Minimum speed of the planet is = Vp = 3.6 * 10¹° meters/sec
                         = 3.6 * 10⁷ kilometers/sec = 36,000,000 km/s
                          36 million km per sec

Answer 2

Answer:

40km/s

Explanation:

See the attachment

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