What is the final velocity of an asteroid on a collision course with Earth?

In summary, a NASA satellite observed an asteroid on a collision course with Earth with an estimated mass of 5 × 10^9 kg and a velocity of 611 m/s relative to Earth. Assuming no friction with the atmosphere, the asteroid will hit the Earth's surface at a speed of 11200 m/s. The correct equation to calculate this is to use the total energy of the asteroid, which is the sum of its kinetic energy and potential energy.
  • #1
Stryder_SW
23
0

Homework Statement



A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass, based on its size, of 5 × 10^9 kg. It is approaching the Earth on a head-on course with a velocity of 611 m/s relative to the Earth and is now 5.0 × 10^6 km away. With what speed will it hit the Earth's surface, neglecting friction with the atmosphere?

Homework Equations



.5(m*v1^2) + mgh = .5(m*v2^2) + mgh

Fg =G(m*me/re^2) where me = mass of Earth and re = radius of Earth or in this case distance of the object. I believe the standard excepted value for mass of the Earth is 5.89*10^24 kg. *edit* and the excepted value for G = 6.67*10^-11 *edit*

The Attempt at a Solution



First off I've converted the distance into meters.
The approach I've been trying is to find the kinetic energy at a speed of 611 m/s (incredibly easy) and find the potential energy at a distance of 5*10^9 m away. Then add them together and solve for v2 in the first equation. Unfortunately the gravitational pull at this distance is not 9.8 m/s. Because of this I've been trying to use the second equation to determine the gravitational pull at that distance and then substitute it into the first equation as g. But try as I might I can not get the right freaking answer. Any help would be greatly appreciated.
 
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  • #3
Unfortunately that gives me a final velocity that's smaller than the initial, which is definitely not correct. Then again I can't rule out the possibility of me doing something incredibly stupid XD.
 
  • #4
Stryder_SW said:
Unfortunately that gives me a final velocity that's smaller than the initial, which is definitely not correct. Then again I can't rule out the possibility of me doing something incredibly stupid XD.

Check your signs. That (-) is there for a reason.
 
  • #5
If I'm not mistaken the (-) sign is only relevant if I have the potential energy on the opposite side from the initial velocity. in my case I do it as follows.

(.5*(5*10^9)*611^2) + ((6.67*10^-11))*(5*10^24)*(5*10^9))/(5*10^9) = (.5*(5*10^9)v2^2) + 0

(9.333025*10^14) + (3.335*10^14) = (2.5*10^9)v2^2

v2 = 711.8433817 m/s

And estimating off an answer to the same problem with a slightly different initial velocity the answer should be somewhere in the 900 to 1000 range
 
Last edited:
  • #6
Your answer looks ok.

Discarding m from all the terms, and using the Standard gravitational parameter μ as 4*105 (in km)

Vi2 + 2*μ *(1/6.73 - 1/5*106) = Vf2

6112 + 126,000 = Vf2 = 500,000

Vf = 707
 
  • #7
Well unfortunately the assignment is past due already, but on the plus side I know the correct answer. apparently its 11200 m/s. not sure how but that's what it says in the answer key.
 
  • #8
LowlyPion said:
Your answer looks ok.
No, it doesn't.

Discarding m from all the terms, and using the Standard gravitational parameter μ as 4*105 (in km)
The Earth's standard gravitational parameter is 398,600 km3/sec2.

Vi2 + 2*μ *(1/6.73 - 1/5*106) = Vf2
The Earth's radius is 6738 kilometers, not 6.73 kilometers.

6112 + 126,000 = Vf2 = 500,000
The initial velocity is 611 m/sec, not 611 km/sec.

Vf = 707
Which is completely wrong.
Stryder_SW said:
If I'm not mistaken the (-) sign is only relevant if I have the potential energy on the opposite side from the initial velocity.
That minus sign is very important.

(.5*(5*10^9)*611^2) + ((6.67*10^-11))*(5*10^24)*(5*10^9))/(5*10^9) = (.5*(5*10^9)v2^2) + 0
This is wrong for several reasons. First off, there is the sign error. Secondly, the asteroid will hit the Earth (6378 kilometers). A lesser error: The Earth's mass is 5.97×1024 kg.

t is a good idea to always carry the units with the numbers. This is part of what got Lowly Pion in trouble.

The total energy of the asteroid is

[tex]E_{tot} = KE + PE = \frac 1 2 m_a v^2 - \frac {GM_e m_a}{r}[/tex]

This is a conserved quantity, so this expression describes the asteroids total energy initially and at the moment it hits the Earth.
 
  • #9
Stryder_SW said:
Well unfortunately the assignment is past due already, but on the plus side I know the correct answer. apparently its 11200 m/s. not sure how but that's what it says in the answer key.

That is the correct answer.
 
  • #10
Oops. Right you are D H. Thanks.

I scratched it out too quickly and dropped the units - twice.

Sorry for any confusion. My haste made waste.
 

Related to What is the final velocity of an asteroid on a collision course with Earth?

1. What is the final velocity of an asteroid?

The final velocity of an asteroid is the speed at which it will be traveling when it reaches its destination or the end of its trajectory. This velocity is affected by factors such as the asteroid's initial velocity, the gravitational pull of other objects, and any external forces acting on the asteroid.

2. How is the final velocity of an asteroid calculated?

The final velocity of an asteroid can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. However, this calculation can become more complex when taking into account other factors such as the gravitational pull of other objects.

3. Why is the final velocity of an asteroid important?

The final velocity of an asteroid is important because it determines how fast and in what direction the asteroid will be traveling. This information is crucial in determining the potential impact and damage of the asteroid on its target.

4. Can the final velocity of an asteroid change?

Yes, the final velocity of an asteroid can change due to various factors such as gravitational pull, collisions with other objects, and external forces. However, once an asteroid has reached its final destination, its velocity will remain constant unless acted upon by another force.

5. How is the final velocity of an asteroid measured?

The final velocity of an asteroid can be measured using various methods, including radar measurements and observations from telescopes. These measurements can provide valuable information about the asteroid's trajectory and potential impact.

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