Podcast #128: We chat with Kent C Dodds about why he loves React and discuss what life was like in the dark days before Git. Listen now.

 6 added 64 characters in body edited Jan 18 '15 at 10:30 ArtOfCode 9,92244 gold badges3535 silver badges7272 bronze badges 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, $$3 \times 10^5 ms^{-1}$$$$3 \times 10^5 \text{ms}^{-1}$$We can work out how much momentum it has: $$\text{momentum }= \text{mass} \times \text{velocity}$$$$\text{momentum (kgms}^{-1})= \text{mass (kg)} \times \text{velocity (ms}^{-1})$$ $$= 100,000 \times (3 \times 10^5)$$ $$= 3 \times 10^{10} kgms^{-1}$$$$= 3 \times 10^{10} \text{ kgms}^{-1}$$ The Sun has a mass of $$1.989 \times 10^{30} kg$$$$1.989 \times 10^{30} \text{kg}$$. Therefore, if we divide the two we can find the resulting velocity of the Sun after impact: $$\frac{3\times 10^{10}}{1.989\times 10^{30}}$$ $$= 1.508\times 10^{-20} ms^{-1}$$$$= 1.508\times 10^{-20} \text{ ms}^{-1}$$ $$= 0.00000000000000000001508 ms^{-1}$$$$= 0.00000000000000000001508 \text{ ms}^{-1}$$ 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. 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, $$3 \times 10^5 ms^{-1}$$We can work out how much momentum it has: $$\text{momentum }= \text{mass} \times \text{velocity}$$ $$= 100,000 \times (3 \times 10^5)$$ $$= 3 \times 10^{10} kgms^{-1}$$ The Sun has a mass of $$1.989 \times 10^{30} kg$$. Therefore, if we divide the two we can find the resulting velocity of the Sun after impact: $$\frac{3\times 10^{10}}{1.989\times 10^{30}}$$ $$= 1.508\times 10^{-20} ms^{-1}$$ $$= 0.00000000000000000001508 ms^{-1}$$ 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. 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. 5 Rollback to Revision 3 edited Jan 18 '15 at 10:27 ArtOfCode 9,92244 gold badges3535 silver badges7272 bronze badges 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 $$100~\mathrm{tons}$$100 tons in mass. That's $$100,000~\mathrm{kg}$$100,000kg. Let's say it's travelling pretty fast, $$3 \times 10^5~\mathrm{m{\cdot}s^{-1}}$$$$3 \times 10^5 ms^{-1}$$We can work out how much momentum it has: $$\text{momentum }= \text{mass} \times \text{velocity}$$ $$= 100,000 \times (3 \times 10^5)$$ $$= 3 \times 10^{10}~\mathrm{kg{\cdot}m{\cdot}s^{-1}}$$$$= 3 \times 10^{10} kgms^{-1}$$ The Sun has a mass of $$1.989 \times 10^{30}~\mathrm{kg}$$$$1.989 \times 10^{30} kg$$. Therefore, if we divide the two we can find the resulting velocity of the Sun after impact: $$\frac{3\times 10^{10}}{1.989\times 10^{30}}$$ $$= 1.508\times 10^{-20}~\mathrm{m{\cdot}s^{-1}}$$$$= 1.508\times 10^{-20} ms^{-1}$$ $$= 0.00000000000000000001508~\mathrm{m{\cdot}s^{-1}}$$$$= 0.00000000000000000001508 ms^{-1}$$ 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. 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 $$100~\mathrm{tons}$$ in mass. That's $$100,000~\mathrm{kg}$$. Let's say it's travelling pretty fast, $$3 \times 10^5~\mathrm{m{\cdot}s^{-1}}$$We can work out how much momentum it has: $$\text{momentum }= \text{mass} \times \text{velocity}$$ $$= 100,000 \times (3 \times 10^5)$$ $$= 3 \times 10^{10}~\mathrm{kg{\cdot}m{\cdot}s^{-1}}$$ The Sun has a mass of $$1.989 \times 10^{30}~\mathrm{kg}$$. Therefore, if we divide the two we can find the resulting velocity of the Sun after impact: $$\frac{3\times 10^{10}}{1.989\times 10^{30}}$$ $$= 1.508\times 10^{-20}~\mathrm{m{\cdot}s^{-1}}$$ $$= 0.00000000000000000001508~\mathrm{m{\cdot}s^{-1}}$$ 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. 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, $$3 \times 10^5 ms^{-1}$$We can work out how much momentum it has: $$\text{momentum }= \text{mass} \times \text{velocity}$$ $$= 100,000 \times (3 \times 10^5)$$ $$= 3 \times 10^{10} kgms^{-1}$$ The Sun has a mass of $$1.989 \times 10^{30} kg$$. Therefore, if we divide the two we can find the resulting velocity of the Sun after impact: $$\frac{3\times 10^{10}}{1.989\times 10^{30}}$$ $$= 1.508\times 10^{-20} ms^{-1}$$ $$= 0.00000000000000000001508 ms^{-1}$$ 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. 4 Adjusted unit formatting edited Jan 18 '15 at 3:10 smithkm 4,4101111 silver badges3333 bronze badges 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$$100~\mathrm{tons}$$ in mass. That's 100,000kg$$100,000~\mathrm{kg}$$. Let's say it's travelling pretty fast, $$3 \times 10^5 ms^{-1}$$$$3 \times 10^5~\mathrm{m{\cdot}s^{-1}}$$We can work out how much momentum it has: $$\text{momentum }= \text{mass} \times \text{velocity}$$ $$= 100,000 \times (3 \times 10^5)$$ $$= 3 \times 10^{10} kgms^{-1}$$$$= 3 \times 10^{10}~\mathrm{kg{\cdot}m{\cdot}s^{-1}}$$ The Sun has a mass of $$1.989 \times 10^{30} kg$$$$1.989 \times 10^{30}~\mathrm{kg}$$. Therefore, if we divide the two we can find the resulting velocity of the Sun after impact: $$\frac{3\times 10^{10}}{1.989\times 10^{30}}$$ $$= 1.508\times 10^{-20} ms^{-1}$$$$= 1.508\times 10^{-20}~\mathrm{m{\cdot}s^{-1}}$$ $$= 0.00000000000000000001508 ms^{-1}$$$$= 0.00000000000000000001508~\mathrm{m{\cdot}s^{-1}}$$ 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. 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, $$3 \times 10^5 ms^{-1}$$We can work out how much momentum it has: $$\text{momentum }= \text{mass} \times \text{velocity}$$ $$= 100,000 \times (3 \times 10^5)$$ $$= 3 \times 10^{10} kgms^{-1}$$ The Sun has a mass of $$1.989 \times 10^{30} kg$$. Therefore, if we divide the two we can find the resulting velocity of the Sun after impact: $$\frac{3\times 10^{10}}{1.989\times 10^{30}}$$ $$= 1.508\times 10^{-20} ms^{-1}$$ $$= 0.00000000000000000001508 ms^{-1}$$ 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. 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 $$100~\mathrm{tons}$$ in mass. That's $$100,000~\mathrm{kg}$$. Let's say it's travelling pretty fast, $$3 \times 10^5~\mathrm{m{\cdot}s^{-1}}$$We can work out how much momentum it has: $$\text{momentum }= \text{mass} \times \text{velocity}$$ $$= 100,000 \times (3 \times 10^5)$$ $$= 3 \times 10^{10}~\mathrm{kg{\cdot}m{\cdot}s^{-1}}$$ The Sun has a mass of $$1.989 \times 10^{30}~\mathrm{kg}$$. Therefore, if we divide the two we can find the resulting velocity of the Sun after impact: $$\frac{3\times 10^{10}}{1.989\times 10^{30}}$$ $$= 1.508\times 10^{-20}~\mathrm{m{\cdot}s^{-1}}$$ $$= 0.00000000000000000001508~\mathrm{m{\cdot}s^{-1}}$$ 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. 3 added 451 characters in body edited Jan 16 '15 at 21:29 ArtOfCode 9,92244 gold badges3535 silver badges7272 bronze badges 2 added 122 characters in body edited Jan 16 '15 at 10:40 ArtOfCode 9,92244 gold badges3535 silver badges7272 bronze badges 1 answered Jan 16 '15 at 10:11 ArtOfCode 9,92244 gold badges3535 silver badges7272 bronze badges