Answer:
a) U = - G m₁m₂ / r , b) K = ½ m (v₀² + 2gy)² or K = 2 mg² y² c) Em = m g y (2 g y + 1)
Explanation:
Let's write the functions that are requested
a) Gravitational power energy
U = - dF / dr
F = G m₁ m₂ / r²
U = - G m₁m₂ / r
r is the height
b) The scientific enrgia
K = ½ m v²
Cinematic
v² = v₀² + 2 g y
K = ½ m (v₀² + 2gy)²
If the initial velocity is zero, the ball is released
K = ½ m 4 g² y²
K = 2 mg² y²
c) Mechanical energy
Em = K + U
Em = 2 m g² y² + m g y
Em = m g y (2 g y + 1)
