Question

Zero, a hypothetical planet, has a mass of 6.0 x 10²³ kg, a radius of 3.4 x 10⁶ m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 10⁷ J, what will be its kinetic energy when it is 4.0 x 10⁶ m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.0 x 10⁶ m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

Answer 1

a)

Total energy (surface) = Total energy (far away point)
- GMm/R + KE (i) = - GMm/r + KE (far)
r = 4.0 *10^6 m
R = 3.4 *10^6 m
KE (far) = KE (i) - GMm[1/R - 1/r]
KE (far) = 5.0 *10^7 - GMm*10^-6[1/3.4 - 1/4]
KE (far) = 5.0 *10^7 - [GMm / 22.66]*10^-6
KE (far) = 5.0 *10^7 - [6.672*10^-11 *6*10²³kg*10kg / 22.66]*10^-6
KE (far) = 5.0 *10^7 - 1.766*10^7
KE (far) = 3.233*10^7 Joule

b)
- GMm/R + KE (Launch) = - GMm/r + KE (far)
KE (launch) = GMm/R - GMm/r + KE (far)

Now

KE (far) = 0.5 m v^2(far)
v^2 (far) = orbital speed in (r) radius = GM/r
KE (far) = GMm/2r

KE (launch) = GMm/R - GMm/r + GMm/2r
KE (launch) = GMm/R - GMm/2r
KE (launch) = GMm*10^-6[1/3.4 - 1/16]
KE (launch) = [ GMm *0.231]*10^-6
= [6.672*10^-11*6*10²³kg*10kg*10^-6 * 0.231]
KE (launch) = 9.27*10⁷ J