# Answer to Question #7862 in Mechanics | Relativity for bhawana

Question #7862
At perihelion on feb 9,1986,Halley&#039;s comet was 8.79*10^7 km from the sun and was moving at a speed of 54.6 km/s relative to the sun.calculate its speed (a)when the comet was 1.16*10^8 km from the sun and (b) at its next apehelion in the year 2024,when the comet will be 5.28*1069 km from the sun.
1
2012-03-30T06:55:33-0400

Perihelion = 8.79 &times; 10^7 km
Aphelion = 5.28 &times; 10^9 km
Generally
speed X distance (to Sun) = constant
So, (speed1) / (speed2) = (distance2) / (distance1)
=&gt; (speed1) = (speed2) X (distance2) / (distance1)
where &#039;2&#039; refers to the known speed and distance and (speed1) is sought for (given) distance1
a) speed1 = 54.6 km/s X (8.79 &times; 10^7) km / (1.16 &times; 10^8) km
= 41.3736 km/s.
b) At, Aphelion = 5.28 &times; 10^9 km
speed1 = 54.6 km/s X (8.79 &times; 10^7) km / (5.28 &times; 10^9) km
= 0.909 km/s
= 41.3736 km/s.

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