Question

You are a visitor aboard the New International Space Station, which is in a circular orbit around the Earth with an orbital speed of vo = 1.10 km/s. The station is equipped with a High Velocity Projectile Launcher, which can be used to launch small projectiles in various directions at high speeds. Most of the time, the projectiles either enter new orbits around the Earth or else eventually fall down and hit the Earth. However, as you know from your physics courses at the Academy, projectiles launched with a great enough initial speed can travel away from the Earth indefinitely, always slowing down but never falling back to Earth. With what minimum total speed, relative to the Earth, would projectiles need to be launched from the station in order to \"escape\" in this way?

Answer 1

Answer:

The minimum total speed is 11.2km/s

Explanation:

We are been asked to find the escape velocity.

Escape velocity is defined as the minimum initial velocity that will take a body(projectile)away above the surface of a planet(earth) when it's projected vertically upwards.

The formula to calculate the escape velocity is Ve = √2gR

For the earth g = 9.8m/s2 , R = 6.4*10^6

Substituting into the equation Ve = √2*9.8*6.4*10^6 = 11.2*10^3m/s

=11.2km/s

Answer 2

Answer:

escape velocity comes out to be = $11kms^{-1}$

Explanation:

The escape velocity corresponds to the initial kinetic energy

Initial kinetic energy = $\frac{1}{2} mv^{2}$

Also we know that work done in lifting the body from earth surface to an infinite distance is equal to the increase in its potential energy  

Increase in P.E = 0 - $(-G\frac{Mm}{R} )=G\frac{Mm}{R}$

M= mass of earth  

R=radius of earth  

now

$\frac{1}{2} mv^{2}$ = G\frac{Mm}{R}[/tex]

$V_{escape}$ = $\sqrt{\frac{2GM}{R} }$

As g = $\frac{GM}{R^{2} }$

$V_{escape}$ = $\sqrt{2gR}$

escape velocity can be calculated as by putting value of  

g = $9.8ms^{-2}$

R = radius of earth = 6400 km  

escape velocity comes out to be = $11kms^{-1}$