An asteroid for birthday

What's with this bugger?

NEO jpl

Set the table settings to
future within a year
Nominal distance less than 1 LD

Only one pops up

50meter in diameter, 7 m/s velocity, and 0.19 lunar distance minimum?

It seems slow, close and quite big will the earth gravity catch it or not?

What do the more savvy people think about this?

Sincerely NC

a reply to: NoConspiracy

It seems slow, close and quite big will the earth gravity catch it or not?

Did you know that Earth's gravity is included in the orbit calculations?

a reply to: Phage

No I have absolutely no clue about the science behind it, thats why i asked.

Thanks alot.

Will it be visible to the naked eye?

I have read that earth has already trapped some rocks that still circle us. Do you know the stats for them?

Sincerely NC

a reply to: NoConspiracy

Will it be visible to the naked eye?

As in most cases, no.

I have read that earth has already trapped some rocks that still circle us. Do you know the stats for them?

No. But it happens. Rocks get caught in Earth's orbit for a while then go on their way.

a reply to: Phage

Do they include the moons gravity as well?

How long will it take for this one to pass between earth and moon?

Like how long from minus -1LD to +1LD

Could the moon push or pull it while he orbits earth and this one pases by?

Sincerely NC

a reply to: NoConspiracy

The asteroid orbits the sun. No idea whether it will pass between earth and moon, probably not.

The orbit calculations use rahter sofisticated models which include relevant forces (sun, jupiter, earth, moon, other plaents, etc). The prediction will be ultimatively limited by the observational data precision though.

a reply to: NoConspiracy

1) Yes.

2) I don't know. I could find out, but it doesn't really matter.

3) Ditto.

4) Yes.

a reply to: moebius

7m/s is very slow, and space very big

384402000 m is one lunar distance.

I'm not sure if i got the math right, but it seems easy.

Divide 2LD by velocity you get seconds, divide by 60 you get minutes, divide by 60 you get hours, divide by 24 you get days divide by 28 the days a lunar orbit around the earth takes and you get the amount of orbits the moon does while he passes by...

2LD ÷ 7 ÷ 60÷60÷24÷28
=
45,4 rotations of the moon arround the earth from the moment he starts to be in between moon and earth, so he surely will be between moon and earth at least 45 times...

This can't be right. Right?

Where did i went wrong?

Sincerely NC

a reply to: NoConspiracy

7m/s is very slow, and space very big

Look again at the units used. You are off by a factor of 1,000.

a reply to: Phage

Thanks for not calling it out in the OP...

embarrassing...

15 hours it will take...

Enjoy your Toilet paper soups

Sincerely NC