Answer:
a) the speed of the two bodies increases as they descend v=√2gy
b) the speed decreases with the time v= v₀ - gt
c) g = G Me/Re² = 9.80 m/s²
d) the acceleration of gravity and is a vector directed towards the center of the Earth
Explanation:
The interaction between bodies and Earth is reflected in the acceleration that bodies acquire in their descent, which is described by the law of universal gravitation as the force in Newton's second law
F = G m Me / r²
F = m a
where m is the mass of the body, Me is the mass of the Earth and r is the distance between them, we substitute
G Me / r² = a
This value of acceleration is called the acceleration due to gravity (a = g) and it is the acceleration that bodies acquire near the Earth.
a) To find the speed of the bodies we can use the kinematic relations
v² = v₀² + 2 gy
suppose bodies start from rest
v= √2 gy
therefore we see that the speed of the two bodies increases as they descend, in the same proportion, except for the effect of the air resistance force that is more important for the lighter bodies.
b) in the case of being launched, the speed decreases with the time
v = v₀ - g t
the decrease is the same for both bodies
c) the acceleration for the two bodies has the same value and is the so-called acceleration of gravity
g = G Me /Re² ( 1+ y/Re)2
Re >>h
g = G Me/Re² = 9.80 m/s²
d) this acceleration is called the acceleration of gravity and is a vector directed towards the center of the Earth
