I have a question vaguely related to the solar system

[Yung Neckpain]

If an asteroid 2158 miles wide (twice the radius of the moon) with twice the density of the moon (6.68 grams per cubic centimeter) were to come close to hitting earth perfectly parallel to earth's orbit around the sun near the northern hemisphere while it's facing directly opposite to the sun at mach 7 (seven times the speed of sound) but miss by 119450 miles (half the distance from the moon to the earth), exactly how much gravitational influence would that asteroid have on the earth's orbit around the sun and what would happen in terms of tidal movements on earth?

Edited December 6, 2017 by Yung Neckpain

[mathematic]

This question belongs in classical physics.

[Yung Neckpain]

My bad, I'll try again in classical physics. Thank you.

[pavelcherepan]

2 hours ago, Yung Neckpain said:

If an asteroid 2158 miles wide (twice the radius of the moon) with twice the density of the moon (6.68 grams per cubic centimeter) were to come close to hitting earth perfectly parallel to earth's orbit around the sun near the northern hemisphere while it's facing directly opposite to the sun at mach 7 (seven times the speed of sound) but miss by 119450 miles (half the distance from the moon to the earth), exactly how much gravitational influence would that asteroid have on the earth's orbit around the sun and what would happen in terms of tidal movements on earth?

Well, this is a rather unlikely scenario. Basically, that asteroid would have to appear out of thin air because there's no way it can exist in such an orbital configuration in the present day Solar System. Such density is also very unlikely to exist given it's size. 

Anyway, simple answer to your question is that it will result in a change in eccentricity of Earth's orbit. The "asteroid" itself would quickly end up on a highly eccentric orbit which would be in some form of resonance with Earth. The mass of this asteroid is still ~10 times less than that of the Earth so while the change in orbit will be noticeable, it won't be enough to completely de-stabilize our orbit.

With tidal movements - we'll experience some quite catastrophic tidal effects. Given the object is around 16 times more massive than the Moon and will pass at half the distance from us, the force exerted should be around 64 times greater than that of the Moon gravity influence. So, it might actually result in some enormous tidal wave, but it won't last for an awfully long time. On the other hand since you specified that it should happen over the North hemisphere, it's also very likely that during the closest approach it will be over land most of the time and then tidal effects will be much less destructive. 

Edited December 6, 2017 by pavelcherepan

[Yung Neckpain]

Thank you, this was extremely helpful

[Janus]

To give an good answer to your question, a few more details are needed.  You say that the asteroid passes on a path that is parallel to the Earth's orbit.  Did you mean that this was the direction of the path at the moment of closest approach, or was this the path the asteroid was following before being effected by the Earth's gravity.  An object with the stated velocity and distance from the Earth at perigee, would have had its trajectory altered by almost 60 degrees from the time it started to be majorly effected by the Earth's gravity and the time it reached perigee.   Given your numbers, I figure that the Earth's orbit would be altered by ~440 m/s.  

Exactly what effect that would have on the Earth's orbit depends on the trajectory of the asteroid. For example, if the trajectory is such that in passing over the Northern hemisphere, the plane of the Earth relative trajectory is at right angles to the ecliptic, then the majority of the change will be to the inclination of the Earth's orbit (~0.84 degrees), with just a small change to the Earth's orbital speed (~3 meter/sec), Again, here we are assuming the the asteroid's path at perigee is parallel to the Earth's orbital path.

As far as tidal effects go, tidal force decreases by the cube of the distance, so, 16 times the mass at 1/2 the distance equals 128 times the tidal effect at perigee.

Over the course of ~6 1/2 days, the tidal force from this body will increase from equal to that of the Moon to 128 times that of the Moon, and then decrease back to equal to the Moon in another 6 1/2 days.

 

[pavelcherepan]

2 hours ago, Janus said:

As far as tidal effects go, tidal force decreases by the cube of the distance, so, 16 times the mass at 1/2 the distance equals 128 times the tidal effect at perigee.

Dang! Thanks for correction.

[Yung Neckpain]

It's the direction of the path at the moment of closest approach

My bad