Question

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surface, the ship's orbital speed is 5500 m/s . By observing the planet, you determine its radius to be 4.48×10^6 m. You then land on the surface and, at a place where the ground is level, launch a small projectile with initial speed 13.6 m/s at an angle of 30.8∘ above the horizontal. If resistance due to the planet's atmosphere is negligible, what is the horizontal range of the projectile? Please explain process

Answer 1

Gravitational constant = G = 6.674 x 10^-11 Nm²/kg²

Mass of the planet = M

Mass of the spaceship = m

Radius of the planet = R_p = 4.48 x 10⁶ m

Altitude of the spaceship above the planet's surface = H = 630 km = 0.63 x 10⁶ m

Radius of the orbit = R

R = R_p + H

R = 4.48x10⁶ + 0.63x10⁶

R = 5.11 x 10⁶ m

Orbital speed of the spaceship = V = 5500 m/s

The centripetal force needed for the circular motion of the spaceship is provided by the gravitational force between the planet and the spaceship.

M = 2.316 x 10²⁴ kg

Gravitational acceleration on the surface of the planet = g

g = 7.701 m/s²

Initial speed at which the projectile is launched = V₀ = 13.6 m/s

Angle at which the projectile is launched = = 30.8^o

Range of the projectile = R

Range of the projectile is given by,

R = 21.127 m

Horizontal range of the projectile = R = 21.127 m