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)
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?
4 answers
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.
(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.
I am not sure we can make those assumptions.
5 answers