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 o=2.45 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 eventually fall down and hit the Earth. However, as you know from your physics courses at the Academy, projectiles launched with a sufficiently great 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? For reference, recall that the radius of the Earth is E=6370000 m
, the mass of the Earth is E=5.98×1024 kg
, the acceleration due to gravity on the surface of the Earth is =9.81 m/s2
and the universal gravitational constant is =6.67×10−11 N·m2/kg2
.

Answer 1

To calculate the minimum total speed required for the projectile to escape the Earth's gravitational pull, we can use the equation for escape velocity:

v_escape = sqrt(2GM/R)

where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.

Plugging in the given values, we get:

v_escape = sqrt(26.67e-115.98e24/6370000)

v_escape = 11186.4 m/s

This is the minimum total speed required for the projectile to escape the Earth's gravitational pull. In order to achieve this speed, the projectile would need to be launched with a velocity of 11186.4 m/s relative to the Earth.

Step-by-step explanation: