Question

a) A space vehicle is launched vertically upward from the Earth's surface with an initial speed of 8.90 km/s, which is less than the escape speed of 11.2 km/s. What maximum height does it attain? Your response differs from the correct answer by more than 10%. Double check your calculations. m (b) A meteoroid falls toward the Earth. It is essentially at rest with respect to the Earth when it is at a height of 2.35 107 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? m/s

Answer 1

Answer:

A) 4037.2[km]; B) 21472[m/s]

Explanation:

A)

This part can be solved using the principle of energy conservation, where kinetic energy will be equal to potential energy. We will define the kinetic energy and the potential energy.

$Ek= 0.5*m*v^2\\where:\\m = mass [kg]\\v = 8.90*10^3[m/s]\\Ep=m*g*h\\where:\\g = gravity = 9.81[m/s^2]\\h = elevation or height [m]$

Now we have to match both equations in this way the mass value is canceled and we can clear h.

$0.5*m*(8.9*10^3)^{2}=m*9.81*h\\ h=4037206.93[m] = 4037.2[km]$

B)

To solve this problem we can use the kinematic equations, but first we must identify the initial data:

yo = 2.35*10^7[m]

y = 0 [m] when hit the ground [m]

vo = 0 [m/s], It is essentially at rest...

$v_{f} ^{2} =v_{o} ^{2} +2*a*(y)\\v_{f} =\sqrt{2*9.81*2.35*10^{7} } \\v_{f} =21472.54[m/s] o 21.47[km/s]$