Calculate the centripetal force exerted on the Earth by the Sun if the period of revolution for Earth is..?

Question

365.25 days, the avg distance is 1.5 x 10^ 8 km and the Earth's mass is 6 x 10^24kg

Answer 1

The centripetal force of the Earth is

m r w^2

where m is the mass of the Earth, r the distance from Earth to Sun, and w the Earth's angular velocity (which is 2 pi / ( 365.25 )( 24 )( 60 )( 60 )) radians / second.

Answer 2

Force = mass x acceleration
acceleration= velocity / time
velocity= distance / time
Force= ?
mass= 6 x 10^24 kg
distance= 1.5 x 10 ^8 km
= 1.5 x 10 ^5 m (meters)
time= 365.25 days
= 31557600 s (seconds)
you have to convert to meters and second because the S.I units for distance is meters and for time is seconds

therefore velocity = (1.5 x 10^5m) / 31557600s
= 4.7 x 10^-3 m/s

and acceleration, a = 4.7 x 10^-3 m/s / 31557600s
= 1.506 x 10^-10m/s^2
And finally force, f= ma
= 6 x 10^24 kg x 1.506 x 10^-10m/s^2
= 9.037 x 10^15 N

Hope that helps

Answer 3

F(centripetal) = mv^2 / r -(1)
is the average distance the circumference for the round trip?
or is it the distance from sun to earth
Im assuming that it is the average distance from sun to earth during rotation.
To aid calculations we will convert (1) into terms of angular velocity
F(c) = m x w^2 x r
(derived from v = rw, so F(c) = m(rw)^2 / r
F(c) = m(w^2 x r^2) / r = m x w^2 x r
w = 2pi / (365.25 x 24 x 3600) in radians per second

once you have this, multiply it by the radius of rotation (avg distance) and the mass of the earth.
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If i got your wording wrong and the distance given was the circumference for rotation,
then velocity v = avg distance/time
and radius is found by
2pi x r = circumference
r = circ / 2pi

radius of rotation