We release an object of mass M at zero speed at a distance of 2*10^6 Km from the center of the earth. What is the speed at which the object hits the earth?

Question

G=6.67*10^-11; g= 9.8 m/s; M=5.97*10^24;
r=6.3781*10^6
Suppose the mass of the object is M= 500 kg. What is the kinetic energy of this object when it hits the earth?
Thanks for your help.

Answer 1

M= 5.97•10²⁴ kg
m=500 kg
r = 6.378•10⁶ m.
R= 2• 0⁶ km= 2• 10⁹ m

GmM/r-GmM/R = mv²/2
GM(1/r -1/R) = v²/2
v=sqrt{2GM(r-R)/rR}=
=sqrt {2•6.67•10⁻¹¹• 5.97•10²⁴(2• 10⁹ - 6.3781•10⁶)/2• 10⁹•6.3781•10⁶}=
=1.1•10⁴ m/s
mv²/2 = 500•(1.1•10⁴)²/23.025 •10¹º J

Answer 2

Thanks Elena for your help. I have been having trouble with some of these equations.