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?
This could only be an MITx problem
I left there in 1959 so this will only be a rough attempt.
NOTE - I am using h, the height for above surface and d for total distance from center. We want f(h) first
Moon center at C
d = r + h distance from center of moon at P
when at P, how much moon do you see?
draw tangents from P to moon circle. Tangents hit at right angle to r thereat A.
at distance d, what is angle PCA?
tan PCA = r/d
so the whole angle at the center C that cuts off your visible sector is
angle = 2 * tan^-1 (r/d)
Now we need to find out how much of the surface area of a sphere is cut out by a central angle.
Well, I know that the area of a zone of a sphere is
A = 2 pi r *height of sector
our height of sector is
r - r tan PCA = r (1 -tan PCA)
so the area of our sector is
A = 2 pi r [r(1-tan PCA) ]
= 2 pi r^2 (1-tan PCA)
But we know tan PCA = r/d
so
Area seen = 2 pi r^2(1-r/d)
as a rough check we can see that when r = d, at the surface, we can not see any thing and at d gets big we see half the sphere.
so now we have the physics problem
A = 2 pi r^2 ( 1 - r/(r+h) )
dA/dt = dA/dh * dh/dt
the physics is about dh/dt which is v
F = G Mm m/d^2
potential energy at distance d (take zero at infinity)
= - G Mm m/d
= - G Mm m/(r+h)^2
kinetic energy = 0 at height h = Hi
so
(1/2)mv^2= G Mm m [1/(r+h)^2 -1/(r+Hi)^2]
or
v^2 = 2 G Mm [1/(r+h)^2 -1/(r+Hi)^2]
and v = dh/dt =2 G Mm [1/(r+h)^2 -1/(r+Hi)^2]^.5
MULTIPLY that by dA/dh
remember A = 2 pi r^2 ( 1 - r/(r+h) )
so
dA/dh = 2 pi r^3 /(r+h)^2
so in the end
dA/dt = 2 pi r^3 /(r+h)^2 *2 G Mm [1/(r+h)^2 -1/(r+Hi)^2]^.5
Now you will have to go back and get h(t)
The satellite can see to the horizon. So along the surface, let the horizon be a distance r from the upcoming inmpact point. Then in basic geometry, one can reduce this to
(d+y)^2=y^2+r^2 (with sherical trig, it is more complicated, but it at small d reduces to the simple plane case)
or d^2+y^2+2dy=y^2 + r^2
or r^2=d^2+2dy
Now the area of what can be seen is approximately PI r^2
or Area=PI r^2
D Area/dt=PI*2r dr/dt
We can find dr/dt from the first equation:
2r dr/dt=2d Dd/dt+2y Dd/dt (I hate folks who use d for distance)
or 2r Dr/dt=(2d+2y) Dd/dt
or d Area/dt=PI *2*(d+y) d'
now what is d', the velocity.
I assume you can get the acceleration of gravity on the Moon. You can first use it as a constant, or figure it as a acceleration as a funciton of distance from the moon. In either case, you then can figure velocity as a function of time, and you have your solution.
I just saw Prof Damon's response. I had the same thought, it was CalTech, Georgia Tech, or MIT. Review his outline first.
if you are using our moon and are close to it, g = (1/6) g earth approximately
I am not sure we can make those assumptions.
To find the rate at which the area of the portion of the moon visible to the satellite decreases, we can consider the change in area over a small time interval "dt" and then take the limit as dt approaches 0.
Let's start by visualizing the situation. As the satellite free falls towards the moon's surface, it moves closer to the moon. The portion of the moon visible to the satellite will decrease as it gets closer to the surface.
To find the rate of change of the visible area, we need to relate the rate of change of the distance of the satellite from the moon's surface (d) to the rate of change of the visible area (A).
We can assume that the satellite remains in the same line of sight as it free falls towards the moon's surface. Therefore, the distance of the satellite from the center of the moon remains constant, given by (r + d).
Now, let's consider a small change in time, dt. During this time, the satellite moves a small distance dx towards the moon's surface, reducing the distance from the surface from (r + d) to (r + d - dx).
The change in area, dA, over this small distance can be approximated as a right triangle with base dx and height (r + d). Therefore, the change in area is given by:
dA = (1/2) * dx * (r + d)
Now, we need to find dx in terms of dt to relate the rate of change of the area (dA/dt) to the rate of change of the distance (ddt).
Since the satellite is free falling, it moves with constant acceleration due to gravity. The equation of motion relating the distance (d) and the time (t) is given by:
d = (1/2) * g * t^2
Differentiating this equation with respect to time, we get:
ddt = g * t
Now, let's relate dx and dt in terms of d and t. Since the satellite is moving closer to the moon's surface, we have dx = -ddt.
Substituting for dd, we get:
dx = -g * t * dt
Now, substituting the value of dx back into the equation for dA:
dA = (1/2) * (-g * t * dt) * (r + d)
Simplifying this equation, we have:
dA = -(1/2) * g * t * dt * (r + d)
Finally, we can find dA/dt by taking the derivative with respect to time:
dA/dt = -(1/2) * g * (r + d)
To find when this rate is maximum, we need to take the derivative of dA/dt with respect to d and set it equal to zero:
d(dA/dt)/dd = -(1/2) * g
Since this is a negative constant, dA/dt is decreasing with respect to d. Therefore, it reaches its maximum value when d is at its minimum, which is when the satellite is closest to the moon's surface.
In summary, the rate at which the area of the portion of the moon visible to the satellite decreases is given by dA/dt = -(1/2) * g * (r + d). This rate is maximum when the satellite is closest to the moon's surface.