A 0.1134kg spherical meteor of diameter 3 cm enters the earth’s atmosphere at an elevation of 121920m with a velocity of 13411.2 m/s. The meteor’s temperature is -73.15˚C. The specific heat at constant volume (Cv) of the meteor is 800 J/kg˚C.
(a) If the meteor has a temperature of 500˚C just before it hits the earth’s surface, what
would its velocity be? Assume that there is no heat transfer to or from the atmosphere
and no work done on or by the atmosphere (due to air drag) on the meteor.
(b) If instead you assumed that the meteor experienced air drag, what would its terminal velocity be in the lower atmosphere? [Thermal conductivity of air = 0.026 W/m˚C]
(c) If just above the earth’s surface the meteor had the terminal velocity determined in part (b) and had a temperature of 25˚C, how much work was done on or by the atmosphere? Is the work positive or negative? Again assume no heat transfer to or from the atmosphere.
To solve these problems, we would need to apply the laws of thermodynamics and some concepts related to motion and drag forces. The following explanations will guide you through the steps required to obtain the answers.
(a) To determine the velocity of the meteor just before it hits the Earth's surface, we need to use the principle of conservation of energy. The initial energy of the meteor is given by its kinetic energy at an elevation of 121,920 m, which can be calculated using the equation:
E_initial = (1/2) * m * v_initial^2
Where m is the mass of the meteor (0.1134 kg) and v_initial is its initial velocity (13,411.2 m/s).
The final energy of the meteor can be calculated considering its temperature change. We can determine the change in internal energy using the specific heat at constant volume (Cv) and the temperature change (ΔT):
ΔU = m * Cv * ΔT
Where m is the mass of the meteor (0.1134 kg), Cv is the specific heat at constant volume (800 J/kg˚C), and ΔT is the change in temperature (500 - (-73.15) = 573.15 ˚C).
Since there is no heat transfer to or from the atmosphere and no work done on or by the atmosphere, the change in internal energy is equal to the change in kinetic energy:
ΔU = ΔKE
Equating the two expressions for ΔU, we have:
m * Cv * ΔT = (1/2) * m * v_final^2 - (1/2) * m * v_initial^2
Cancelling out the mass, we get:
Cv * ΔT = (1/2) * v_final^2 - (1/2) * v_initial^2
Solving for v_final, we have:
v_final^2 = (2 * Cv * ΔT) + v_initial^2
Substituting the known values, we can calculate v_final.
(b) To determine the terminal velocity of the meteor in the lower atmosphere, we need to consider the effects of air drag. The drag force experienced by an object moving through a fluid can be expressed as:
F_drag = 0.5 * ρ * A * Cd * v^2
Where ρ is the density of the fluid (air in this case), A is the cross-sectional area of the object (π * r^2 for a spherical object), Cd is the drag coefficient, and v is the velocity of the object.
The terminal velocity is reached when the drag force equals the weight of the object:
F_drag = F_weight
Equating the two expressions, we have:
0.5 * ρ * A * Cd * v_terminal^2 = m * g
Where m is the mass of the meteor (0.1134 kg) and g is the acceleration due to gravity (9.8 m/s^2).
Simplifying, we get:
v_terminal^2 = (2 * m * g) / (ρ * A * Cd)
To find the terminal velocity, we need the values of ρ, A, and Cd. Given that the diameter of the sphere is 3 cm, we can calculate the cross-sectional area (A = π * r^2). ρ is the density of air, which is approximately 1.225 kg/m^3. The drag coefficient, Cd, depends on the shape of the object.
(c) To calculate the work done on or by the atmosphere, we need to consider the work done by the drag force (W_drag). The work done by a force is given by the equation:
W_drag = ∫F_drag * ds
Where F_drag is the drag force and ds is the displacement along the path of motion. In this case, we need to consider the displacement from just above the Earth's surface to the final position.
The work done by the drag force can be calculated using the formula:
W_drag = F_drag * d
Where F_drag is the drag force, and d is the distance traveled.
Since the meteor has reached its terminal velocity, its acceleration is zero:
F_drag = F_weight
Using this relationship, we can substitute the values of F_drag and d into the equation to calculate the work done on or by the atmosphere.
Remember to always double-check your calculations and units to ensure accurate answers.
To solve the given problem, we need to apply the principles of conservation of energy and Newton's laws of motion. Let's solve part (a) first:
(a) To find the final velocity of the meteor, we can use the principle of conservation of energy. The initial kinetic energy of the meteor is given by:
KE_initial = (1/2) * m * v_initial^2
Given:
Mass of the meteor (m) = 0.1134 kg
Initial velocity (v_initial) = 13411.2 m/s
Next, we need to find the final kinetic energy of the meteor using its final temperature. The specific heat at constant volume (Cv) is given as 800 J/kg˚C. Let's denote the change in temperature as ΔT:
ΔT = final temperature - initial temperature
= (500 - (-73.15)) ˚C
= 573.15 ˚C
The change in internal energy (ΔU) of the meteor can be calculated as:
ΔU = Cv * m * ΔT
Finally, we can calculate the final kinetic energy of the meteor using the principle of conservation of energy:
KE_final = KE_initial + ΔU
= (1/2) * m * v^2 + Cv * m * ΔT
To find the final velocity (v), we rearrange the equation:
v = sqrt((KE_final - Cv * m * ΔT) / (1/2 * m))
Substituting the given values:
v = sqrt(((1/2) * m * v_initial^2 + Cv * m * ΔT - Cv * m * ΔT) / (1/2 * m))
= sqrt(v_initial^2)
= v_initial
Therefore, the velocity of the meteor just before hitting the earth's surface would still be 13411.2 m/s.
(b) To determine the terminal velocity of the meteor considering air drag, we can use the concept of drag force. The drag force (Fd) acting on the meteor can be calculated using the equation:
Fd = (1/2) * Cd * ρ * A * V^2
Where:
Cd is the drag coefficient (dimensionless), which depends on the shape of the object
ρ is the density of air (kg/m^3)
A is the cross-sectional area of the meteor (m^2)
V is the velocity of the meteor with respect to the air (m/s)
First, let's find the drag coefficient (Cd). Since the meteor is spherical, the drag coefficient can be approximated using Stokes' law:
Cd = (24 / Re) + (2.6 * Re / (24 + Re))
Where Re is the Reynolds number, given by:
Re = (ρ * V * D) / μ
D is the diameter of the meteor (m)
μ is the dynamic viscosity of air (Pa*s)
Given:
Diameter of the meteor (D) = 0.03 m
Density of air (ρ) = 1.225 kg/m^3
Dynamic viscosity of air (μ) = estimate as 1.8 * 10^-5 Pa*s (for temperatures around 25˚C)
Plugging in the values, we can calculate Cd and then find the terminal velocity (V_terminal) by balancing the drag force with the gravitational force acting on the meteor:
Fd = mg
where m is the mass of the meteor, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
(c) To calculate the work done on or by the atmosphere when the meteor is at its terminal velocity just above the earth's surface, we can use the work-energy principle. The work done (W) is given by:
W = ΔKE
where ΔKE is the change in kinetic energy of the meteor as it comes to a stop. Since the meteor has reached its terminal velocity, the change in kinetic energy is zero, so the work done is also zero. Therefore, no work is done on or by the atmosphere in this situation. The work is neither positive nor negative.
Note: The initial and final temperatures of the meteor are not directly used to calculate the velocity, terminal velocity, or work done. They are given as additional information in the problem statement.