A satellite is free falling from a distance (d) of the surface of the moon with radius (R), with zero initial velocity. At what rate does the area of the portion of the moon visible to the satellite decrease at a time (t) after the free fall starts? When is this rate maximum? Explore your answer by assigning numerical values of your choice to the solution
MITx question ?
y=√x
Do you have the radius of the moon?
Do you have the formula for the distance of the horizon from a height h?
Do you have the gravitational acceleration g for the moon?
Steve-that is all the given info
Damon- no.
thanks Eryk i will try it
it is an application of derivative
Is this question Solved.. If Yes please send me at vikasmonga95(at)gmail(dot)com
To determine the rate at which the area of the portion of the moon visible to the satellite decreases, we can start by visualizing the situation.
Since the satellite is free-falling, it will follow a parabolic path toward the surface of the moon. As it falls, the portion of the moon visible to the satellite will decrease. We need to find the rate of change of this area with respect to time (t).
Let's assume that the distance of the satellite from the surface of the moon (d) is greater than the radius of the moon (R). We'll assign some numerical values to help illustrate the concept:
d = 100 km
R = 50 km
To calculate the area of the portion of the moon visible to the satellite, we can first find the angle of the portion of the moon from the satellite's perspective. This angle can be determined using trigonometry.
θ = arctan(R / (R + d))
Now, we can determine the area of the portion of the moon visible to the satellite at time (t):
A = π * R^2 * (2π - 2θ)
To find the rate at which this area decreases with respect to time (t), we can differentiate A with respect to t:
dA/dt = d(π * R^2 * (2π - 2θ))/dt
Using the chain rule, we can break down this expression into:
dA/dt = dA/dθ * dθ/dt
The first derivative dA/dθ can be calculated as:
dA/dθ = -2π * R^2
The second derivative dθ/dt represents the rate at which the angle θ changes with respect to time (t). This can be determined by considering the motion of the satellite.
Since the satellite is free-falling, it experiences constant acceleration toward the moon's surface. We can use the equations of motion to find the second derivative:
dθ/dt = d(dθ/dt)/dt
Using the equation of motion, we can express the second derivative as:
dθ/dt = -g / (R + d)
Where g represents the acceleration due to gravity.
Now, we can substitute the values we assigned to d and R into our equations to find the rate at which the area of the portion of the moon visible to the satellite decreases.
dA/dt = -2π * R^2 * (-g / (R + d))
Based on this expression, we can observe that the rate at which the area decreases is proportional to the gravitational acceleration (g) and inversely proportional to the distance (d + R) between the satellite and the moon's surface. This makes sense since the closer the satellite gets to the moon, the faster the visible area will decrease.
To find when this rate (dA/dt) is maximum, we can set the expression for dA/dt equal to zero and solve for time (t). Since dA/dt is proportional to g and inversely proportional to (d + R), the maximum rate occurs when the satellite is closest to the moon's surface.
I hope this explanation helps! Let me know if you have any further questions.