A space vehicle is launched vertically upward from the Earth

A space vehicle is launched vertically upward from the Earth\'s surface with an initial speed of 7.21 km/s, which is less than the escape speed of 11.2 km/s. What maximum height does it attain? A meteoroid falls toward the Earth. It is essentially at rest with respect to the Earth when it is at a height of 2.31 Times 10^7 m above the Earth\'s surface. With what speed does the meteorite (a meteoroid that survives to impact the Earth\'s surface) strike the Earth?

Solution

a) speed of space vehicle achieved , v = 7.21 km/s

v = 7210 m/s

R = 6371 km

R = 6.371 *10^6 m

M = 5.98 *10^24 Kg

let the maximum height is h

Using conservation of energy

0.5 * m* v^2 - G * m * M/R = - G * m * M/(R + h)

0.5 * v^2 - * M/R = - G* M/(R + h)

0.5 * 7210^2 - 6.673 *10^-11 * 5.98 *10^24/(6.371 *10^6) = -6.673 *10^-11 * 5.98 *10^24/(6.371 *10^6 + h )

solving for h

h = 1.896 *10^6 m

the maximum height attained by the space vehicle is 1.896 *10^6 m

b)

height , h = 2.31 *10^7 m

let the speed at the surface of earth is v m/s

Using conseravtion of energy

0.5 * m* v^2 - G * m * M/R = - G * m * M/(R + h)

0.5 * v^2 - * M/R = - G* M/(R + h)

0.5 * v^2 - 6.673 *10^-11 * 5.98 *10^24/(6.371 *10^6) = -6.673 *10^-11 * 5.98 *10^24/(6.371 *10^6 + 2.31 *10^7 )

solving for v

v = 10167 m/s

the speed of meteriot when it strike the earth is 10167 m/s