Question

A person of mass 75 kg is standing on the surface of the Earth.

a. Calculate the total energy of this individual, assuming they are at rest relative to the surface. (Hint: Use the expression for gravitational potential energy (*NOT U_g=mgh*)). Although the person is standing at rest relative to the surface of the Earth, they are rotating along with the Earth and thus store some kinetic energy.) Ignore the rotation of the Earth about the Sun, the motion of the solar system in the galaxy, and the motion of the galaxy in the universe.

b. How much energy would need to be added for this individual to go into circular orbit a negligible height above the surface of the Earth?

c. How many times would this individual orbit the Earth every day?

Answer 1

(A) PE = - G M m / R

PE = - (6.67 x 10^-11)(5.976 x 10^24)(75)/(6371 x 10^3)

PE = - 4.69 x 10^9 J

and KE = m v^2 / 2

and v = w R = (2 pi / T)(R)

KE =(75 / 2)(2 pi / (24 x 3600 s))^2 (6371 x 10^3)^2

KE = 8.05 x 10^6 J  

TE = PE + KE = - 4.68 x 10^-9 J  

(B) for circular path,

Fg = m a_c

G M m / R^2 = m v^2 / R

m v^2 = G M m / R

TE = PE + KE =- G M m / R + m v^2 / 2 = - G M m / R + G M m/ 2 R

TE = - G M m / 2 R

= - 2.345 x 10^9 J

Energy needed = TEf - TEi = 2.33 x 10^9 J  

(C) G M / R = v^2

and f = v / (2 pi R)

f = sqrt(6.67x 10^-11 x 5.976 x 10^24 / (6371 x 10^3)) / (2 x pi x 6371 x 10^3)

f = 1.97 x 10^-4 s^-1

in a day, n = (24 x 3600)(1.97 x 10^-4)

n = 17 times