Question

1.Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed v0(<vesc). Express in terms of v0,rE,ME, and G.

2. How high does a rocket go if 8.60 km/s ? Ignore air resistance and the Earth's rotation.

Answer 1

1. Using the law of conservation of energy, rocket's PE + KE on the earth's surface = PE at the highest point

Or, - GMm/r + (1/2)mv^2 = - GMm/(r + h), [h = height of the rocket, m = mass of rocket]

Or, GM/(r+h) = GM/r - (1/2)v^2

Or, h = GM / [GM/r - (1/2)v^2] - r

Or, h = rv^2 / [2GM/r - v^2]

Or, in the given symbols, formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed v0(<vesc) is h= (rE) * (v0)^2 / [2G(ME)/(rE) - (v0)^2]

2. Now, using the values, mE = 5.972 * 10^24, rE = 6.371 * 10^6 m, G = 6.67 x 10^-11 , v= 8.6km/s= 8600m/s

h = { (6.371*10^6) * (8600)^2 } / { 2 * 6.67*10^-11 * 5.972*10^24 / (6.371*10^6) - (8600)^2 }

So, h= (4.71*10^14) / { (1.25*10^8) - (7.396*10^7) } = 9223738m= 9223.74 km