Advanced Engineering Problem: Asteroid on a collision course

In summary: I think you are correct about the 66 years. You should also consider the fact that the asteroid will not be coming straight towards the sun, but at an angle. This will make the distance it travels even greater.In summary, the conversation discusses a question about an asteroid from the Kuiper belt falling towards the Sun and the calculations needed to determine its time and velocity when it reaches Earth's orbit. It involves using the conservation of energy equation and solving a differential equation to find the position and time as a function of each other. The final answer for the time is approximately 66 years with a velocity of 2.2 km/s.
  • #1
Mark Brewer
38
4

Homework Statement



A start to solving this question would be much appreciated.

Question: An asteroid from the Kuiper belt (approximately 30 AU from the Sun) is knocked out of its circular orbit and begins to fall straight inwards towards the Sun. We want to know how long it will take to get here and how fast it will be moving when it does.

Using the conservation of energy we have:

E =1/2(mv^2) - (GMm/r)

where E is the total energy of the system, m is the mass of the asteroid v is its velocity, G is Newton's constant, M is the mass of the Sun, and r is the distance from the asteroid to the Sun. As the asteroid begins its journey very far away (compared to where it ends up) we may approximate its initial potential energy by zero. Also, assuming (rather unrealistically) that the asteroid starts falling from rest, we may take its initial kinetic energy to be zero as well, so that E = 0

a) Write down the differential equation for r(t) giving the position of the asteroid as a function of time. Pay close attention to signs.

b) Solve the differential equation for t(r), the time as a function of position.

c) Plug in the numbers to find out how long it will take the asteroid to reach Earth orbit. Take G = 6.67 x 10^-11 (Nm^2) / (kg^2) and M = 2 x 10^30 kg.

d) How fast will it be moving when it crosses Earth's orbit (in km/s)?

e) Assuming the asteroid has a mass m = 9 x 10^20 kg (approximately equal to that of Ceres), compute the kinetic energy released if it were to collide with the Earth. Remark: For comparison the energy released by the asteroid that wiped out the dinosaurs was estimated to be around 4 x 10^23 J.

------------------------------------------------------------------------

The Attempt at a Solution

My attempt to question #1:

E = 0

r = (v^2) / 2GM (conservation of energy equation)

r = r1 ((v+v1) / 2) t (constant acceleration)

r = ((v^2) / GM(v+v1) (1./t))

dr/dt = ((v^2) / GM(v+v1) (1/t^2))

Am I on the right track?
My attempt to question #2:

dr/dt = ((v^2) / GM(v+v1) (1/t^2))

by separable variables

(t^-2)dt = (GM (v2 + v) / v^2)dr

by integration I get

* = (GM (v2 + v) / v^2) r

t = cube root ((v^2/(GM(v2+v)r) +c)

Am I still on the right track?

My attempt to question #3:

t = 0

I know I'm lost now.
 
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  • #2
[itex] v = \dot{r} [/itex]. The energy conservation equation itself is a differential equation for [itex] r [/itex].
 
  • #3
Daeho Ro said:
[itex] v = \dot{r} [/itex]. The energy conservation equation itself is a differential equation for [itex] r [/itex].

Am I on the right track now?

t - t0 = ∫ (1 / sqrt((2/m)(E-v(x))) dx
 
  • #4
From
[tex] E =\dfrac{1}{2} m v^2 - G\dfrac{Mm}{r} = 0, [/tex]
the velocity is given by
[tex] v = \sqrt{\dfrac{2GM}{r}} = \dfrac{dr}{dt}. [/tex]

Now, just integrate it with suitable initial conditions, e.g. when [itex] t=0 [/itex], the radius [itex] r = R (50 \text{AU}) [/itex] .
 
  • #5
Daeho Ro said:
From
[tex] E =\dfrac{1}{2} m v^2 - G\dfrac{Mm}{r} = 0, [/tex]
the velocity is given by
[tex] v = \sqrt{\dfrac{2GM}{r}} = \dfrac{dr}{dt}. [/tex]

Now, just integrate it with suitable initial conditions, e.g. when [itex] t=0 [/itex], the radius [itex] r = R (50 \text{AU}) [/itex] .
where does R = 50 AU come from? The OP mentions 30 AU.
 
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  • #6
SteamKing said:
where does R = 50 AU come from? The OP mentions 30 AU.
You are correct. It's terrible mistake. :wideeyed:
 
  • #7
I'm getting t = 211 years until the asteroid strikes the Earth. With initial conditions, R = 31 AU; t0 = 0

Is this correct?
 
  • #8
Mark Brewer said:
I'm getting t = 211 years until the asteroid strikes the Earth. With initial conditions, R = 31 AU; t0 = 0

Is this correct?
Why did you take R = 31 AU? The OP said the Kuiper Belt is approximately 30 AU from the sun.
 
  • #9
I added an additional AU because the Earth is an AU from the Sun.
 
  • #10
Mark Brewer said:
I added an additional AU because the Earth is an AU from the Sun.
The OP specifically stated the Kuiper Belt was approximately 30 AU from the sun, not from the earth.
 
  • #11
Thank you, SteamKing. I guess I was interested in the impact of Earth. I'll recalculate.
 
  • #12
If I've calculated correctly my t = 2430s or 40.5 min; and the velocity of the asteroid would be 1512 mph.
 
  • #13
Mark Brewer said:
If I've calculated correctly my t = 2430s or 40.5 min; and the velocity of the asteroid would be 1512 mph.
Is this the time it takes the asteroid to fall all the way from the Kuiper Belt to the earth? What happened to 211 years?
 
  • #14
Yes. The time from the Kuiper Belt to Earth's orbit. I'm still asking myself the same question about 211 years. Twice I've calculated r = 30AU and I determined t = 2430 s. I'll crunch the numbers again and see if I can derive a t of approx. 211 years.
 
  • #15
I have to get ready for class. I'll recalculate several times for t. My first answer for t is now 65.9 years.
 
  • #16
Mark Brewer said:
Yes. The time from the Kuiper Belt to Earth's orbit. I'm still asking myself the same question about 211 years. Twice I've calculated r = 30AU and I determined t = 2430 s. I'll crunch the numbers again and see if I can derive a t of approx. 211 years.
2430 sec. is only about 40 minutes. If an object took that little time to travel 30 AU, it would be zipping by Earth at considerably more than 1500 mph.

Remember, it takes light about 8 minutes to travel 1 AU from the sun to the earth.
 
  • #17
Thank you, Daeho Ro and Steamking. My final answer for t is 66.1 years, v is 2.2 km/s, K = 2.18x10^21 J.

Mark
 

Related to Advanced Engineering Problem: Asteroid on a collision course

1. What are the potential consequences of an asteroid on a collision course with Earth?

If an asteroid were to collide with Earth, it could have catastrophic consequences. The impact would generate a powerful shockwave, causing widespread destruction and potentially triggering earthquakes and tsunamis. The explosion could also release a large amount of dust and debris into the atmosphere, blocking out the sun and causing a cooling effect on the planet. Depending on the size of the asteroid, it could potentially lead to mass extinction events.

2. How likely is it for an asteroid to be on a collision course with Earth?

The chances of an asteroid colliding with Earth are very low, as the vast majority of asteroids in our solar system are located in the asteroid belt between Mars and Jupiter. However, there is still a small possibility that an asteroid could be on a collision course with Earth at any given time. Scientists are constantly monitoring the skies for any potential threats and have developed methods for diverting an asteroid's path if necessary.

3. What measures can be taken to prevent an asteroid collision?

If an asteroid is identified as being on a collision course with Earth, there are several methods that can be used to divert its path. These include using a spacecraft to either directly push the asteroid off course or to use a gravitational tractor to gradually alter its trajectory. Another option is to detonate a nuclear device near the asteroid to change its direction. However, these methods would need to be implemented well in advance to be effective.

4. How do scientists track and monitor potentially hazardous asteroids?

Scientists use a variety of methods to track and monitor potentially hazardous asteroids. This includes ground-based telescopes, space-based telescopes, and radar systems. Ground-based telescopes are used to scan the skies and identify any new objects, while space-based telescopes can track objects over longer periods of time. Radar systems can also provide more precise information on an asteroid's size, shape, and trajectory.

5. What is being done to prepare for a potential asteroid impact?

NASA and other space agencies have developed plans and strategies for dealing with potential asteroid impacts. This includes ongoing monitoring and tracking of near-Earth objects, as well as research and development of methods for deflecting or destroying an asteroid. There are also international efforts to improve our understanding of the threat and to coordinate response efforts in case of an actual impact.

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