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
Answer:
B. 175 N
Explanation:
Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.
Mathematically, net force is given by the formula;
Where;
Fnet is the net force
Fapp is the applied force
Fg is the force due to gravitation
In this scenario, we observed that both forces are acting in the same direction.
Therefore:
Net force = 100 N + 75 N
Net force = 175 Newton
Net force would be towards the right and back (opposite direction of motion) since it's slowing down (decelerating) and turning right.
Answer:
h=17357.9m
Explanation:
The atmospheric pressure is just related to the weight of an arbitrary column of gas in the atmosphere above a given area. So, if you are higher in the atmosphere less gass will be over you, which means you are bearing less gas and the pressure is less.
To calculate this, you need to use the barometric formula:
Where R is the gas constant, M the molar mass of the gas, g the acceleration of gravity, T the temperature and h the height.
Furthermore, the specific gas constant is defined by:
Therefore yo can write the barometric formula as:
at the surface of the planet (h =0) the pressure is ![P_0[\tex]. The pressure at the height requested is half of that:[tex]P=\frac{P_0}{2}](https://tex.z-dn.net/?f=P_0%5B%5Ctex%5D.%20The%20pressure%20at%20the%20height%20requested%20is%20half%20of%20that%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DP%3D%5Cfrac%7BP_0%7D%7B2%7D)
applying to the previuos equation:
solving for h:
h=17357.9m
