For this problem, take the distance from the Earth to the Sun to be 1.5×10 11 m

Question

For this problem, take the distance from the Earth to the Sun to be 1.5×1011m and the radius of the Earth to be 6.38×106m.

A) Using the fact that the earth orbits the Sun once per year, calculate the Earth's orbital angular speed (in rad/s) due to its motion around the sun.   

B) Using the fact that the Earth orbits around its axis once per day, calculate the earth's angular speed (in rad/s) about its axis. (Assume the Earth rotates at a constant rate).   

C) Calculate the tangential speed of the Earth around the Sun (assuming a circular orbit).   

D) Calculate the tangential speed of a point on the Earth's equator due to the planet's axial spin.

E) Calculate the magnitude of the radial acceleration of the point in part D.

F) Calculate the magnitude of the tangential acceleration of the point in part D.

Explanation / Answer

here,

distance from earth to sun , x = 1.5 * 10^11 m

radius of earth , r = 6.38 * 10^6 m

A)

time period , T = 365 days

T = 31536000 s

the angular speed , w = 2*pi /T

w = 2*pi/31536000

w = 2 * 10^-7 rad/s

B)

time period , = 1 days

T = 86400 s

the angular speed , w = 2*pi /T

w = 2*pi/86400

w = 7.27 * 10^-5 rad/s

C)

tangential speed of the Earth around the Sun , v = ( r + x) * w

v = ( 1.5 * 10^11 + 6.38 * 10^6 ) *2 * 10^-7

v = 3 * 10^4 m/s

the tangential speed of the Earth around the Sun is 3 * 10^4 m/s

D)

the tangential speed of a point on the Earth's equator due to the planet's axial spin , v = r * w

v = 6.38 * 10^6 * 7.27 * 10^-5

v = 463.83 m/s

the tangential speed of a point on the Earth's equator due to the planet's axial spin is 463.83 m/s

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