Neutron star: free fall acceleration The textbook from which I teach physics at the end of secondary school, has a question about a neutron star: $M_{star}=1.4\cdot M_{sun}$, radius 15km. "Calculate the free fall acceleration at the surface of the neutron star". Pupils are supposed to use $a=F_g/m=G*M_{star}/R^2$


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*Is the free fall acceleration the same as the coordinate acceleration for a hypothetical observer at rest on the star surface?

*Is the free fall acceleration the same as the coordinate acceleration for an observer at rest at a great distance from the star?

*Does the free fall acceleration at the surface have the same value according to both observers?

*Is the Newtonian approach $a=F_g/m=G*M_{star}/R^2$ correct, considering the strong gravity at the surface?
 A: I'm guessing your questions all amount to whether general relativistic effects become important at the surface of a neutron star. To answer this we can compare the flat space metric (in polar coordinates):
$$ ds^2 = -c^2dt^2 + dr^2 + r^2 d\Omega^2 \tag{1} $$
with the Schwarzschild metric that describes the geometry outside a spherically symmetric mass:
$$ ds^2 = -\left(1-\frac{2GM}{c^2r}\right)c^2dt^2 + \frac{dr^2}{\left(1-\frac{2GM}{c^2r}\right)} + r^2 d\Omega^2 \tag{2} $$
The difference is that factor of $1-2GM/c^2r$, which we can also write as $1-r_s/r$ where $r_s$ is the Schwarzschild radius - $r_s = 2GM/c^2$. Feeding in the mass and radius of the neutron star we find this factor is about $0.72$, so general relativistic effects are indeed important.
Your question (1) is answered in What is the weight equation through general relativity?. The coordinate acceleration measured by an observer at the surface is:
$$ a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} \tag{3} $$
so it differs from the Newtonian prediction by (in this case) a factor of about $\sqrt{0.72}$.
Re your questions (2) and (3), offhand I don't know the expression for the coordinate acceleration measured far from the star, but it will not be the same as equation (3). A distant observer sees falling objects slow as they approach the event horizon and asymptotically approach zero speed at the horizon. So the coordinate acceleration is obviously different from the coordinate acceleration measured near the horizon.
A: Having read John Rennie's answer above, I'm going to give an answer that's hopefully of the same sense, which hopefully makes sense, but which hopefully brings out an  issue.     
1. Is the free fall acceleration the same as the coordinate acceleration for a hypothetical observer at rest on the star surface?
Yes and no. Yes because the falling body falls the way it falls, regardless of how the two observers might describe it. However it's no because the observer on the surface is subject to gravitational time dilation. He's going slower, so he would say things fall faster. 
2. Is the free fall acceleration the same as the coordinate acceleration for an observer at rest at a great distance from the star?
Yes, and no. Yes because the body falls the way it falls, and because this question is looking for a Newtonian answer wherein the free-fall acceleration is that described by the distant observer using a simple expression. However it's no because the Newtonian answer is starting to diverge from the GR answer, and there's a contradiction looming.  
3. Does the free fall acceleration at the surface have the same value according to both observers?
No. Their seconds are different, they don't agree that it's circa 8.2 x 10¹¹ m/s². And even when they compare notes and make allowance for the gravitational time dilation, that contradiction is still looming. LOL, it reminds me of the end of The Terminator when the boy says there's a storm coming.  
4. Is the Newtonian approach $a=F_g/m=G*M_{star}/R^2$ correct, considering the strong gravity at the surface?
No. Now take a look at what John Rennie said when he referred to the black hole. The distant observer sees the falling object slow to zero speed. So in the limit, the distant observer doesn't say gravity is strong at the surface. He says it goes away. 
Houston we have a problem!   
