Question

A meteoroid passes through a position in space where its speed is very small relative to Earth's and it is at a perpendicular distance of 19 Earth radii above Earth's surface. The meteoroid is moving in such a way that Earth captures it.
What is the speed of the meteoroid when it is one Earth radius above the ground

Answer 1

Answer:

$v=7506.4m/s$

Explanation:

If we call 1 the position in space where the meteoroid of mass m speed is very small relative to Earth's (whose mass is $M_E=5.98\times10^{24}kg$ and radius is $R_E=6371000m$) and it is at a perpendicular distance of $h_1=19R_E$ above Earth's surface, and 2 the position when it is $h_2=R_E$ above the ground, then, since in this case mechanical energy is conserved, we can write:

$E_1=E_2$

which means:

$K_1+U_1=K_2+U_2$

where K is the kinetic energy and U the gravitational potential energy. Since $K_1=0J$ we can write:

$\frac{-GM_Em}{r_1}=\frac{mv_2^2}{2}+\frac{-GM_Em}{r_2}$

which means:

$v_2=\sqrt{2GM_E(\frac{1}{r_2}-\frac{1}{r_1})}$

And since $r=R_E+h$ (the distance of an object to the center of the Earth is Earth's radius plus the height of the object), we have:

$v_2=\sqrt{2GM_E(\frac{1}{R_E+h_2}-\frac{1}{R_E+h_1})}=\sqrt{2GM_E(\frac{1}{2R_E}-\frac{1}{20R_E})}=\sqrt{\frac{GM_E}{R_E}(1-\frac{1}{10})}=\sqrt{\frac{0.9GM_E}{R_E}}$

This for our values is:

$v_2=\sqrt{\frac{0.9GM_E}{R_E}}=\sqrt{\frac{0.9(6.67\times10^{-11}Nm^2/kg^2)(5.98\times10^{24}kg)}{6371000m}}=7506.4m/s$