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

v = 7506.4 m/s

Answer 2

`v=7506.4m/s`

Step-by-step 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*10^(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:

`(-GM_Em)/(r_1)=(mv_2^2)/(2)+(-GM_Em)/(r_2)`

which means:

`v_2=\sqrt{2GM_E((1)/(r_2)-(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((1)/(R_E+h_2)-(1)/(R_E+h_1))}=\sqrt{2GM_E((1)/(2R_E)-(1)/(20R_E))}=\sqrt{(GM_E)/(R_E)(1-(1)/(10))}=\sqrt{(0.9GM_E)/(R_E)}`

This for our values is:

`v_2=\sqrt{(0.9GM_E)/(R_E)}=\sqrt{(0.9(6.67*10^(-11)Nm^2/kg^2)(5.98*10^(24)kg))/(6371000m)}=7506.4m/s`