If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how far from the

If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how far from the Earth's center is it when its speed is 1/7 of the escape speed.

I only have two more tires for this question, please help me.

When the PE = mGM/r^2 * r = 1/2 mV^2 = KE we find the escape velocity as V = sqrt(2GM/r). Then v = sqrt(2GM/R) = V/7 where r = Earth's radius, M is Earth's mass, and G is G. R = ? is the distance you are looking fr.

So V^2 = 2GM/r and v^2 = 2GM/R = (V^2)/49. So we have (v/V)^2 = (V^2/49)/V^2 = 1/49 = 2GM/R//2GM/r and r/R = 1/49 so that R = 49r. ANS.

The velocity v = V/7 when the projectile is 49 Earth radii from the center of the Earth. ANS.