Answer:
Q = 1057.5 [cal]
Explanation:
In order to solve this problem, we must use the following equation of thermal energy.
where:
Q = heat energy [cal]
Cp = specific heat = 0.47 [cal/g*°C]
T_final = final temperature = 32 [°C]
T_initial = initial temperature = 27 [°C]
m = mass of the substance = 450 [g]
Now replacing:
Answer:
a) K = 2/3 π G m ρ R₁³ / R₂ , b) U = - G m M / r
Explanation:
The law of universal gravitation is
F = G m M / r²
Part A
Let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / R₂
G m M / R₂² = m v² / R₂
v² = G M / R₂
They give us the density of the planet
ρ = M / V
V = 4/3 π R₁³
M = ρ V
M = ρ 4/3 π R₁³
v² = 4/3 π G ρ R₁³ / R₂
K = ½ m v²
K = ½ m (4/3 π G ρ R₁³ / R₂)
K = 2/3 π G m ρ R₁³ / R₂
Part B
Potential energy and strength are related
F = - dU / dr
∫ dU = - ∫ F. dr
The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1
U- U₀ = G m M ∫ dr / r²
U - U₀ = G m M (- r⁻¹)
We evaluate for
U - U₀ = -G m M (1 / - 1 /
)
They indicate that for ri = ∞ U₀ = 0
U = - G m M / r