(a) In Newton’s time the radius of the Earth (6371 km) and the acceleration due

Question

(a) In Newton’s time the radius of the Earth (6371 km) and the acceleration due to gravity on the Earth’s surface (9.81 m/s 2 ) were known. Use this information to find ke for the Earth. Use MKS units.

(b) Find the orbital radius of the moon. Write the answer in miles. Hint. Assume the moon’s orbit is circular. From class r 2 = ke /r 2 where ke was calculated in (a).

(c) Find the radius of a geosynchronous orbit (so the orbital period about the Earth is 24 hours).

K is a constant

e : eccentricity

Explanation / Answer

acceleration due to gravity,

g = G M / r^2 = ke / r^2

ke = (9.81 m/2^2) ( 6371 x 10^3 m)^2 = 3.98 x 10^14 m^3 / s^2

(B) for moon's orbit,

G M m / r^2 = m w^2 r

G M / r^2 = w^2 r

and G M = ke = 3.98 x 10^14 m^3 / s^2

w = 2pi / T = 2pi / (27 day x 24 hr/day x 3600 s/hr) = 2.69 x 10^-6 rad/s
  
3.98 x 10^14 = (2.69 x 10^-6)^2 ( r^3 )

r = 380 x 10^6 m Or 236158 miles

(C) w = 2pi / T = 7.27 x 10^-5 rad/s

3.98 x 10^14 = (7.27 x 10^-5)^2 (r^3)

r = 422 x 10^5 m

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.