Object falling towards moon

That will be the distance it falls in time t.In summary, an object is dropped from a distance of 2 times the moon's radius from the moon's center. It will fall a distance of 1.74*10^6 meters with a varying acceleration due to the moon's gravity. The speed at impact will be 1678.54 m/s. Using the equation a = v*dv/dx, the velocity can be expressed as a function of distance and then integrated to find the distance as a function of time. This gives the time it takes for the object to fall and the distance it falls in that time.
  • #1
dbogen
2
0

Homework Statement


An object is dropped from a distanse of 2*R(moon radius) from moon senter.
How many seconds does it take until impact with moon, and in what speed will it hit?

The distanse of the fall will be 1.74*10^6 m. The problem is the not so constant accleration.. ;)

gravity constant: G := 6.67*10^(-11)
moon mass M := 0.0735^24 kg
moon radius R := 1.74*10^6 m

Homework Equations



gravity(accleration) is
g=G*M/R^2

So a(x)=(G*M)/(2*R - x)^2 , x = meter fallen x{0..R)

The Attempt at a Solution


I've calculated the speed at impact:

Average accleration:
> A:=(G*M)/(2*R - x)^2
> Aa := (int(A, x = 0 .. R))/R;
0.8096264368

Time using x= 0.5*a*t^2
> T := solve(R = 0.5*Aa*t^2, t);
-2073.229031, 2073.229031

This time is found using constant accleration, so it isn't the correct one...
but it will work for finding the speed of impact.

Speed
> V := T*Aa;
-1678.541033, 1678.541033
So it will hit at 1678,54 m/s

but for how long will it fall?
 
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  • #2
You have acceleration as a function of distance. Put it in the expression a = v*dv/dx.That way you can obtain velocity as a function of distance by integrating. That will solve the first part of your problem. For time you can write v as dx/dt. Now if you put that and integrate you have distance as a function of time.
 
  • #3
Thanks aim!

I got the correct velocity using your method, but i don't figure out the distanse as a function of time...

Could you please try to explain it again, a bit more detailed? :)
 
  • #4
Ok you figured out velocity as a function of distance(x). Now what you do is write v as dx/dt. You will notice that the equation now contains x and t only. A little rearrangement and integration and you can express x as function of t.
 
  • #5

Based on the given information, it appears that the object is being dropped from a distance of 2 times the radius of the moon, and the acceleration due to gravity is not constant due to the changing distance from the moon's center. Using the equation for acceleration due to gravity, g = G*M/R^2, we can calculate the average acceleration for the entire fall using the given values for the moon's mass (M) and radius (R). This average acceleration, Aa, is approximately 0.8096264368 m/s^2.

Next, we can use the equation x = 0.5*a*t^2 to find the time (t) it takes for the object to fall a distance of R, which is the radius of the moon. This equation assumes constant acceleration, so it will not give us the exact time it takes for the object to hit the moon's surface, but it will give us an estimate. Solving for t, we get two solutions: -2073.229031 seconds and 2073.229031 seconds. Since time cannot be negative, we can discard the negative solution and conclude that it will take approximately 2073.229031 seconds for the object to fall a distance of R.

Finally, to find the exact speed at which the object will hit the moon's surface, we can use the equation V = T*Aa, where V is the final velocity, T is the time it takes to fall a distance of R, and Aa is the average acceleration we calculated earlier. Plugging in the values, we get two solutions: -1678.541033 m/s and 1678.541033 m/s. Again, we can discard the negative solution and conclude that the object will hit the moon's surface at a speed of approximately 1678.541033 m/s.

However, it is important to note that these calculations are based on the assumption of constant acceleration, which is not the case in this scenario. A more accurate and complex calculation would take into account the changing acceleration as the object falls towards the moon.
 

1. How does the force of gravity affect an object falling towards the moon?

The force of gravity is the same regardless of the object's mass or distance from the moon. This means that all objects will experience the same acceleration (9.8 m/s²) towards the moon due to its gravitational pull.

2. Will the object ever reach the moon's surface?

Yes, if the object is falling towards the moon with enough initial velocity, it will eventually reach the moon's surface. However, if it does not have enough initial velocity, it may enter into a stable orbit around the moon or continue to fall towards the moon at a slower rate.

3. How does air resistance affect an object falling towards the moon?

Air resistance will have a negligible effect on an object falling towards the moon, as there is no atmosphere on the moon to create resistance. However, if the object is passing through Earth's atmosphere on its way to the moon, air resistance will slow it down.

4. What is the rate of acceleration for an object falling towards the moon?

The rate of acceleration for an object falling towards the moon is approximately 1.62 m/s². This is significantly less than the acceleration due to gravity on Earth (9.8 m/s²) due to the moon's smaller mass.

5. How does the moon's gravity compare to Earth's gravity?

The moon's gravity is approximately 1/6th of Earth's gravity. This means that an object falling towards the moon will experience a weaker gravitational force than if it were falling towards Earth. However, the rate of acceleration due to gravity is still the same (9.8 m/s²) on both the moon and Earth.

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