Question

A meteoroid, heading straight for Earth, has a speed of 14.8 km/s relative to the center of Earth as it crosses our moon's orbit, a distance of 3.84 × 108 m from the earth's center. What is the meteroid's speed as it hits the earth? You can neglect the effects of the moon, Earth's atmosphere, and any motion of the earth.

Answer 1

Answer:

Explanation:

M = Mass of Earth = 5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

$v_i$ = Launch velocity = 14.8 km/s

$v_f$ = Final velocity

r = Orbit distance = $3.84\times 10^8\ m$

m = Mass of satellite

As the energy of the system is conserved we have

$U_i+K_i=U_f+K_f\\\Rightarrow -\dfrac{GMm}{r}+\dfrac{1}{2}mv_i^2=-\dfrac{GMm}{R}+\dfrac{1}{2}mv_f^2\\\Rightarrow -\dfrac{GM}{r}+\dfrac{1}{2}v_i^2=-\dfrac{GM}{R}+\dfrac{1}{2}v_f^2\\\Rightarrow \dfrac{1}{2}v_f^2=\dfrac{GM}{R}-\dfrac{GM}{r}+\dfrac{1}{2}v_i^2\\\Rightarrow v_f=\sqrt{2GM(\dfrac{1}{R}-\dfrac{1}{r})+v_i^2}\\\Rightarrow v_f=\sqrt{2\times 6.67\times 10^{-11}\times 5.972\times 10^{24}\times (\dfrac{1}{6.371\times 10^6}-\dfrac{1}{3.84\times 10^8})+14800^2}\\\Rightarrow v_f=18493.53507\ m/s$

The meteroid's speed as it hits the earth is 18493.53507 m/s