Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

What size asteroid is needed to move the Sun?

+0
−0

What kind of asteroid would it take to hit our sun out of its current position, even by just 50 meters? And how big does it need to be in order for it to do so? Would it continue to travel through space after being knocked out of our system, or is this simply not possible? If so:

What size asteroid would be able to 'shatter' the sun?

History
Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

This post was sourced from https://worldbuilding.stackexchange.com/q/8800. It is licensed under CC BY-SA 3.0.

0 comment threads

1 answer

+0
−0

tl;dr: It can't be done.

Meteors and all the various different classifications of them (meteorites, meteoroids) are small. The sun is big. More importantly, the sun is hot. If a meteor is heading for the sun, not only would it do almost nothing when it hits, it would be vaporised long before it hits.

However, in theory:

Let's assume you have a meteor 100 tons in mass. That's 100,000kg. Let's say it's travelling pretty fast, $3 \times 10^5 \text{ms}^{-1}$We can work out how much momentum it has:

$$ \text{momentum (kgms}^{-1})= \text{mass (kg)} \times \text{velocity (ms}^{-1}) $$ $$ = 100,000 \times (3 \times 10^5) $$ $$ = 3 \times 10^{10} \text{ kgms}^{-1} $$

The Sun has a mass of $1.989 \times 10^{30} \text{kg}$. Therefore, if we divide the two we can find the resulting velocity of the Sun after impact:

$$ \frac{3\times 10^{10}}{1.989\times 10^{30}} $$ $$ = 1.508\times 10^{-20} \text{ ms}^{-1} $$ $$ = 0.00000000000000000001508 \text{ ms}^{-1} $$

It is important to note that this same math applies to a vaporised asteroid as conservation of momentum still applies; however, if some of the vaporised material passes the Sun, it won't affect it. However, as shown in this article, this is unlikely as the acceleration of dispersion is not enough to disperse the material sufficiently before reaching the Sun.

So even a fairly "heavy" meteor only results in the Sun moving at a fraction of a metre per second, at which speed its gravity would keep all the planets in orbit with it.

Lastly, it is impossible to "shatter" the sun, as it is made of gas and would simply part and reform around an asteroid.

History
Why does this post require moderator attention?
You might want to add some details to your flag.

0 comment threads

Sign up to answer this question »