Question

a projectile is launched from the surface of the earth. its initial speed is 46% of the speed required to escape earth. what altitude above earth's surface does it reach, as a multiple of earth's radius re?

Answer 1

The altitude above earth's surface it reaches, as a multiple of earth's radius re is 1.13 R.

How to calculate the altitude above earth's surface?

The altitude above earth's surface it reaches, as a multiple of earth's radius re is calculated as follows;

The formula for escape velocity is given as;

v₀ = √(GM/2R)

We will apply the law of conservation of energy as follows;

¹/₂mv₀² - GMm/R = -GMm/r

¹/₂m(0.46 x GM/2R) - GMm/R = -GMm/r

0.46GMm/4R - GMm/R = - GMm/r

(0.46GMm - 4GMm)/4R = - GMm/r

-3.54GMm/4R = - GMm/r

-3.54/4R = -1/r

-3.54r = -4R

r = 4R/3.54

r = 1.13 R

Thus, the altitude above earth's surface it reaches, as a multiple of earth's radius re is 1.13 R.