Question

A rocket is projected upward from the earth's surface (r = R_E) with an initial speed v₀ that carries it to a distance r = 1.5 R_Efrom the center of the earth. What is the launch speed v₀? Assume that air friction can be ignored.

Answer 1

given that ::

M_e = Mass of earth = 5.972 x 10²⁴ kg

R_e = radius of earth = 6370 km

R_i = R_e

R_f = 1.5 R_e

the launch speed is given as ::

KE = PE                   ( eq. 1 )

(1/2) m v_o² = (G*M m / r) - (G * M m/ (r x 1.5))

(1/2)v_o² = (G *M / r) - (G * M / (r x 1.5)) = PE

v_o = ? (2 * PE)                           ( eq. 2 )

PE = G M [ 1 / r - 1 / 1.5r ]                                          ( eq. 3 )

where, iniversal gravitational constant = 6.67 x 10-11 N m² kg^-2

M = mass of earth

r = radius of earth

PE = (6.67 x 10^-11 N m² kg^-2 ) ( 5.972 x 10²⁴ kg) [ 1 / 6370 - 1.5 x 6370 ]   

PE = 39.8199 x 10¹³ [ 1 / 6370 - 1 / 9555 ]

PE = 39.8199 x 10¹³ [ 1 / 19110 ] = 0.002083 x 10¹³

using eq. 2,

v_o = ? (2 * PE) = ? ( 2 x 0.002083 x 10¹³ ) = ? ( 0.004166 x 10¹³ km )   

v_o = ?4.166 x 10¹³ m/s

v_o = 6.45 x 10⁶ m/s