As mars orbits the sun in its elliptical orbit, its distanceof closest approach

Question

As mars orbits the sun in its elliptical orbit, its distanceof closest approach to the center of the sun is 1.067 X 10^11 m. if the orbital speed of Mars at this point is2.198 X 10 ^4 m/s , what is the orbital speed when it is at adistance of 2.492 X 10 ^11 m from the center of the sun?   As mars orbits the sun in its elliptical orbit, its distanceof closest approach to the center of the sun is 1.067 X 10^11 m. if the orbital speed of Mars at this point is2.198 X 10 ^4 m/s , what is the orbital speed when it is at adistance of 2.492 X 10 ^11 m from the center of the sun?  

Explanation / Answer

We have:     v = rw => w = (v/r)       (1) With v = 2.198 x 104m/s and r = 1.067 x1011m. Substituting the above values in equation (1),we get:      w = [(2.198 x 104)/(1.067x 1011)] => w = 2.0599 x 10-7 s-1 The orbital speed when it is at a distance of 2.492 x1011 m from the center of the sun is        v =rw        (2) With r = 2.492 x 1011 m and w = 2.0599 x10-7 s-1 Substituting the above values in equation (2),we get:       v = 2.492 x 1011 x2.0599 x 10-7 => v = 5.13 x 104 m/s. Substituting the above values in equation (2),we get:       v = 2.492 x 1011 x2.0599 x 10-7 => v = 5.13 x 104 m/s.

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