A projectile is launched vertically from the surface of the earth with an initia

Question

A projectile is launched vertically from the surface of the earth with an initial speedv0 = 6.84 km/s.

(a) What is the magnitude of its acceleration (m/s^2) 1910 km above thesurface of the earth?

(b) At what height (km) will it stop and begin to fall back to the surface of the earth?

G = 6.673x10^-11 m^3 / kg*s^2

Radius of the Earth = 6.37x10^6

Mass of the Earth = 5.97x10^24

The correct answers are (a) 5.80 m/s^2, (b) 3800 km.

I know how to calculate (a). The equation is derived was GM/(Radius of the Earth + height above the Earth)^2

However, I am stuck on (b). I do not know how the answer is 3800 km.

The equation that I think I should use, but am not sure, is.

Uf- Ui= GMEm(1/rf-1/ri)

Can someone please show the work for this problem? Thanks!

Explanation / Answer

b).5*m*v*v-GM*m/(Radius of the Earth + height above the Earth)=0

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