Speed of Impact of meteoroid

[MEwill]

Homework Statement

Q: "Two meteoroids are heading for earth. Their speeds as they cross the moons orbit are 1km/s.
The first meteoroid is heading straight for earth, what is it's speed of impact?"

Velocity = 1km/s
Rearth = 6.37E6 m
Dmoon = 3.84E8 m
G = 6.673E-11 N*m^2/kg^2

Homework Equations

E = 1/2mv$^{2}$ - √G*Mearth*m/r

The Attempt at a Solution

PE = (G * Mearth * m)/Dmoon
KE = -PE
1/2mv$^{2}$ = -PE

From there can cancel the masses (lower-case m) of the meteoroid and solve for v. I don't feel like that is the right track. Shouldn't it be as the PE goes to 0? Any help would be appreciated. Thanks!

 

[gneill]

Homework Statement

Q: "Two meteoroids are heading for earth. Their speeds as they cross the moons orbit are 1km/s.
The first meteoroid is heading straight for earth, what is it's speed of impact?"

Velocity = 1km/s
Rearth = 6.37E6 m
Dmoon = 3.84E8 m
G = 6.673E-11 N*m^2/kg^2

Homework Equations

E = 1/2mv$^{2}$ - √G*Mearth*m/r

The Attempt at a Solution

PE = (G * Mearth * m)/Dmoon
KE = -PE
1/2mv$^{2}$ = -PE

From there can cancel the masses (lower-case m) of the meteoroid and solve for v. I don't feel like that is the right track. Shouldn't it be as the PE goes to 0? Any help would be appreciated. Thanks!

Hi MEwill, Welcome to Physics Forums.

I'm not sure where you got the square root of G in the energy formula in your Relevant Equations, you might want to check that.

The standard zero reference for gravitational potential energy is taken to be at infinite distance, so PE will not be going to zero as the meteoroid approaches the planet's surface.

Instead, determine the total energy E at the given location (just crossing the Moon's orbit) and assume that that energy is conserved, so it's the same when it reaches the planet's surface.

 

[MEwill]

The square root was a typo! I wasn't using it in my calculations. Figured it out I'm pretty sure. I wasn't adding the PE's. It should be the sum of the potential energy when crossing the moons orbit and the potential energy just before is reaches the Earth's surface (using radius of the earth) then set that = to the KE and solve for v? Just to make sure my logic is on the right track.

 

[gneill]

Not sure what you mean by "adding the PE's". You want to employ a conservation law, which in this case is conservation of mechanical energy (KE + PE = constant).

When the body in orbit is of insignificant mass with respect to the primary (here the Earth is the primary massive body) then you can work with what's called "specific" quantities. That's energy quantities given in "per unit mass" amounts. Essentially that means you just disregard the mass of the projectile when writing your conservation of energy equation. This is fine, since its mass always cancels out in the process of solving for the velocity anyways

In this problem you find what's known as the "specific mechanical energy" of the body and use the fact that this quantity is conserved:
$$\xi = \frac{v^2}{2} - \frac{\mu}{r}$$
where $\mu = G M$, and $M$ being the mass of the Earth.

$\xi$ is conserved along the trajectory of the projectile (meteoroid). So write the expression for $\xi$ for the two locations of interest and set them equal...

 

[HalfLostAndSearching]

I would approach this problem by first considering the relevant physical principles involved. In this case, we are dealing with the motion of objects in a gravitational field. From this, we can use the equation for gravitational potential energy:

PE = -G * M * m / r

Where G is the gravitational constant, M is the mass of the larger object (in this case, the Earth), m is the mass of the smaller object (meteoroid), and r is the distance between the two objects.

We can also use the equation for kinetic energy:

KE = 1/2 * m * v^2

Where m is the mass of the object and v is its velocity.

Since we are interested in the speed of impact, we can assume that the meteoroid has reached its maximum velocity at the time of impact. Therefore, we can set the kinetic energy equal to the potential energy at the point of impact:

1/2 * m * v^2 = -G * M * m / r

We can then solve for v:

v = √(2 * G * M / r)

Plugging in the given values for G, M (mass of Earth), and r (radius of Earth + distance to moon), we get a speed of approximately 25.5 km/s.

In conclusion, the speed of impact for the first meteoroid would be approximately 25.5 km/s. It is important to note that this is just a theoretical calculation and actual speeds may vary depending on the specific conditions and factors involved.