Question

Given that the mass of the sun and the earth are respectively 1.99x10^30 kg and 5.98x10^24 kg, and that, the distance between them is 1.50x10^11 m, (a) find the graviational attraction between the sun and the earth. (b) Taking one year to equal 3.156x10^7 s, and that the orbit of the earth around the sun is circular, find, in miles per hour, the speed (considered uniform) of the earth around the sun.

Answer 1

(a) We know that the force due to gravitational attraction is given by
F = GMm/r²
where G is gravitational constant = 6.67*10^-11
M is mass of sun = 1.99*10³⁰ kg
m is mass of earth = 5.98*10²⁴ kg
r is distance between them = 1.5*10¹¹ m
F = (6.67*10^-11 )*( 1.99*10³⁰ *5.98*10²⁴)/(1.5*10¹¹ )² = 3.528*10²² N
(b) We know that
Linear speed is given by (V)= 2r/T
Where T is the time period to complete one revolution = 3.156*10⁷ s
V = (2*1.5*10¹¹ )/(3.156*10⁷ ) = 29863.05 m/s
Now in miles /hour
1 m/s = 2.23694 mi/hr
V =  29863.05*2.23694 = 66801.74 mi/hr