(a) Determine the circumference of the Earth through the equation,
C = 2πr
Substituting the known values,
C = 2π(1.50 x 10¹¹ m)
C = 9.424 x 10¹¹ m
Then, divide the answer by time which is given to a year which is equal to 31536000 s.
orbital speed = (9.424 x 10¹¹ m)/31536000 s
orbital speed = 29883.307 m/s
Hence, the orbital speed of the Earth is ~29883.307 m/s.
(b) The mass of the sun is ~1.9891 x 10³⁰ kg.
Formula for potential energy is V=mgh, where m is mass in KG, g is earth acceleration (10 m/s^2), and h its height in meters. We know mass, acceleration is constant and also known, we know height also. Lets substitute
V=75*10*300=225000[J]=225[kJ] - its the answer
