An object lying on a planet\'s equator is accelerated (a) toward the center of p
ID: 1992765 • Letter: A
Question
An object lying on a planet's equator is accelerated (a) toward the center of planet because the planet rotates with a period of 2.4 day, (b) toward the central star because the planet revolves around the star in an almost circular orbit with a period of 2.1 y, and (c) toward the center of a galaxy because the star moves around the galactic center with a period 1.2 x 108 y. Take the radius of the planet to be 6.2 x 106 m, the radius of the planetary orbit around the star to be 1.1 x 1011 m, and the radius of the orbit around the galactic center to be 2.7 x 1020 m and calculate the magnitudes of these three accelerations as multiples of g = 9.8 m/s2.Explanation / Answer
Only centripetal accelaration is to be calculated which is given by
a=rw^2 where w=angular velocity
So for planet
ap=rp(wp)^2=(6.2x10^6)[2/(2.4x24x3600)]^2 m^2/s=5.692x10^-3 m^2/s=5.81x10^-4 g
for star
as=rs(ws)^2=(1.1x10^11)[2/(2.1x365x24x3600)]^2 m^2/s=1.01x10^-4 g
for galaxy
ag=rg(wg)^2=(2.7x10^20)[2/(1.2x10^8x365x24x3600)]^2 m^2/s=7.59x10^-11 g