A projectile is shot directly away from Earth\'s surface. Neglect the rotation o

Question

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius R_E gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.505 of the escape speed from Earth and (b) its initial kinetic energy is 0.505 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.) (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?

Explanation / Answer

Potential Energy due to earth on any mass at a distance r = -GMm/r

So potential energy on surface = -GMm/R

Escape velocity from earth when it reaches infinite distance where earth has no influence.

Just to escape the earth, the final velocity i.e at infinity = 0

So By conservation of energy -GmM/R + 0.5 m v^2 = 0 +0 (both final potential and kinetic energies are zero)

So v(escape) = sqrt( 2MG/R)

Now given its initial velocity in problem(a) = 0.505 v(escape)

Let the projectile reaches a distance X.

So by conservation -GmM/R + 0.5 m v^2 = - GmM/ X + 0 and v is given as 0.505 v(escape)

So -GmM/R + 0.505^2 (GmM/R) = -GmM/X

So -0.7449GmM/R = -GmM/X

so X = R/0.7449 = 1.342 R

(b) Given its Initial KE = 0.505 * escape KE

Escape KE = GmM/R

So again by conservation of energy -Gmm/R + 0.505 * (-GMm/R) = -GmM/X

So X = R/0.495 = 2.0202 R

Mechanical Energy = Total Energy = potential energy + Kinetic energy.

Since To escape earth Total initial energy = total final energy = 0+0

So we need zero mechanical energy to make a projectile escape.

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