Ahuge cannon is assembled on an airless planet having insignificant axial spin. the planet has a radius of 5.00 × 106 m and a mass of 1.46 × 1023 kg. the cannon fires a projectile straight up at 2000 m/s. an observation satellite orbits the planet at a height of 1000 km. what is the projectile's speed as it passes the satellite? (g= 6.67 × 10-11 n · m2/kg2)

1830 m /s

Explanation:

The potential energy of the projectile at  the surface of planet

- G x 1.46 x 10²³ m  / 5 x 10⁶

Kinetic energy at surface

1/2 m v²

Total energy = - G x 1.46 x 10²³ m  / 5 x 10⁶ + 1/2  m v²

- 6.67 X 10⁻¹¹ x 1.46 x 10²³ /5 x 10⁶  m + 1/2 m x 2000²

- 1.947 x 10⁶ m + 2m x 10⁶

=  .053  x 10⁶ m

Potential energy of projectile at 1000 km height

= - G x 1.46 x 10²³ m  / 6 x 10⁶

= -  1.6225 x 10⁶  m J  

Total energy

= - 1.6225 x 10⁶ + 1/2 m V ²

Applying conservation of energy in gravitational field at surface and height

- 1.6225 x 10⁶  m + 1/2 m V ² = .053  x 10⁶ m

1/2 V ²  = ( .053 + 1.6225 ) x 10⁶

1/2 V² = 1.6755  x 10⁶ m /s

V = 1.83 X 10³ m/s

1830 m /s

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.