Escape velocity

Question

Calculate the escape velocity from a white dwarf and a neutron star. Assume that both the white dwarf and the neutron star is 1 solar mass. Let the white dwarf’s radius be $10^{4}$ kilometres and the neutron star’s radius be 10 kilometres.

escape velocity $= \sqrt{\frac{2GM}{R}}$

'G' is gravitational constant (6.733 minus ¹¹) 'M' is mass 'R' is radius

So I tried inserting the values for the white dwarf where in

R is $10^{4}$ G is $6.67\times10^{-11}$ and M is 1 because 1 solar mass

and the answer is $1.15\times10^{-9}$

I'm pretty not that much sure if I answered this one correctly or not, because I'm in doubt with the 1 solar mass that I've got, do I still need to convert some things in here?

1 solar mass is the mass of our sun ($1.989 \times 10^{30}$kg) not the number 1 — Triatticus, Mar 25 '16 at 15:07
This question is not well worded. For example, "Assume that each (what??) 1 solar mass" and "let the white dwarf's radius be (how many???) kilometers" — Brevan Ellefsen, Mar 25 '16 at 15:10
@BrevanEllefsen sorry just edited it, i have a weak internet connection right now — user325861, Mar 25 '16 at 15:12
@Dan so it would be like this ? $\sqrt{\frac{2(\left ( 6.67x10^{-11} \right ) (1.989×103^{30}))}{10^{4}}}$ — user325861, Mar 25 '16 at 15:15
It seems to me that you made something strange with units. Have you transformed all the data in international units? (meters, kg, ...) — user3209698, Mar 25 '16 at 15:17
@seoanes The units that I've used is still the ones for the G which is N.m^2/kg^2 R which is Km and M which is Kg — user325861, Mar 25 '16 at 15:20
Although now you just have to be careful about the two radii, one is ten thousand km while the other is only 10 — Triatticus, Mar 25 '16 at 15:20
@user325861 the radious has to be written in meters, in order to be consistent with the units of $G$ — user3209698, Mar 25 '16 at 15:21
Yes because the gravitational constant is in Newtons squared meter per square kilogram — Triatticus, Mar 25 '16 at 15:22
so the 10 km would become 10,000 m and the 10^{4} would be 10,000,000 m ? am I correct? — user325861, Mar 25 '16 at 15:24

score 2 · Accepted Answer · edited Mar 25 '16 at 15:53

2

The key issue here is not math, it is units. Your equation for escape velocity seems to be correct, as per Wikipedia.

$$v_\text{escape}=\sqrt{2GM \over R}=\sqrt{{2\left(6.674\times 10^{-11} {\text{m}^3\over{\text{kg}\cdot s^2}}\right)\left(1 M_{\odot} \right)}\over{\left(1\times 10^4 \text{ km}=1\times10^7 \text{ m}\right)}}$$ This converts to: $$3.653\times 10^{-9} \sqrt{\mathbf{{m^3 M_{\odot}}\over{\text{m}\cdot \text{kg} \cdot \text{s}^2}}}$$ This is technically correct, but we can convert solar mass ($M_\odot$) to $\text{kg}$ by the following conversion: $1 M_\odot = 1.989\times 10^{30} \text{ kg}$.

Note that we will multiply the number by the square root of the relationship, since we factored out the square root. So the final answer is: $$5.152\times 10^6 \sqrt{\mathbf{\text{m}^2 \over \text{s}^2}}$$ $$5.152\times 10^{6} \mathbf{m\over{s}}$$

edited Mar 25 '16 at 15:53

Michael Hardy

1

answered Mar 25 '16 at 15:24

FundThmCalculus

2,513

I think the result is of the order of $\sim10^6 m/s$. Are you sure about the substitution of the solar mass? – user3209698 Mar 25 '16 at 15:26
I think you made an error: the proportionality constant between the solar mass and kg should be square rooted, so the result should be on the order of $10^{-9+15}=10^6$. (Possibly $10^7$, since the original number was almost on the order of $10^{-8}$.) – Ian Mar 25 '16 at 15:27
I was fixing that as you guys were commenting. Good catch! I also misinterpreted $10^4$ as m instead of $10^4 km$. That's been fixed too. – FundThmCalculus Mar 25 '16 at 15:27
@FundThemCalculus wait shouldn't be the formula is $\sqrt{\frac{2GM}{R}}$ – user325861 Mar 25 '16 at 15:27
Yup, fixing that too. – FundThmCalculus Mar 25 '16 at 15:28
@FundThmCalculus advance thank you !! – user325861 Mar 25 '16 at 15:28
@FundThmCalculus another question i tried doing it on calculator I got the answer of 5152588.864 m/s – user325861 Mar 25 '16 at 15:38
That's what I got too, so no worries there:$5,152,589=5.153\times 10^6$, which is within roundoff of my original answer. The relative difference is about one thousandth of a percent. – FundThmCalculus Mar 25 '16 at 15:44
@FundThmCalculus oh thank you, so for the second one the neutron star problem i'll just follow the steps for it right? – user325861 Mar 25 '16 at 15:51
Yup, exactly the same problem. – FundThmCalculus Mar 26 '16 at 02:33

Escape velocity

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