A planet travels in an elliptical orbit about a star as shown. At what pair of points is the speed of the planet the same?

Explanation

A Pitot tube (also called Prandt tube or Wind speed indicator) consists of a bent tube ABCD with small apertures at A and D. This tube is inserted in pipe P through which the fluid (air in this case) is flowing. Let ` rho ` be the density of air. The plane of the aperture A is parallel to the direction of air flow. As such, the velocity ` (v_1 ) ` of air at A is the same as its velocity in pipe (which is equal to the velocity v of plane relative to air). Let ` P _ 1 ` be the pressure at A. The plane of the aperture at D is normal to the direction of air-flow in the pipe. The velocity ` (v_2 )` of air-flow rapidly falls to zero at D and the pressure increases to ` P_2 ` . Applying Bernoulli’s theorem to the horizontal flow at A and D, we get
` P_1 +(1)/(2) rho v _ 1^ 2 = P_2 + (1)/(2) rho v_2 ^ 2 or v = sqrt ((2(P_1 - P_2))/(rho))) ` (as ` v_1 = v ` and ` v _ 2 = 0` )
but `P_1 - P_2 = h sigma g ` where ` sigma ` is the density of the liquid in the tube ABCD.
As ` h = 40 cm = 0.4 m, sigma = 0.8 xx 1000 kg// m ^ 3 , rho = 1 kg// m^3 , v = sqrt ((2 xx 0.4 (0.8 xx1000 ) 9.8)/(1)) m//s = 79.2 m//s `