Will the earth spiral into the sun? If not, why not?
No. Because this is not true:
the distant observer measures the earth's orbital velocity to be only half the velocity necessary to keep the earth in stable orbit around the sun
No derivation of this claim is provided. It is simply claimed without any justification. The claim is incorrect.
General relativity is based on tensors and pseudo Riemannian geometry. They are explicitly designed from first principles so that all measurable results are independent of the coordinate system. If the orbit is stable in one frame then it is necessarily stable in all frames. The foundations of the math do not permit any other outcome.
Invariant spacetime curvature is assumed to be the cause of orbit.
As is stated here, curvature is invariant. So the results do not depend on the coordinates.
EDIT 1: I wanted to add the actual GR math. The outcome of this is exactly as expected by anyone with GR experience. Indeed, from first principles it could be no other way. All equations are using geometrized units where $c=G=1$.
We start with the weak field metric in cylindrical coordinates: $$ds^2 = (1 - 2 U) dr^2 + (-1 - 2 U) dt^2 + (1 - 2 U) dz^2 + (r^2 - 2 r^2 U) d\phi^2 $$ with the standard gravitational potential in cylindrical coordinates $$ U=-\frac{M}{\sqrt{r^2+z^2}} $$
Now, an object in orbit is in free-fall, so the worldline of the planet is a geodesic. To calculate the orbit of the earth we therefore calculate the geodesic using the equations described here. When we do so, we get the following equations: $$0 = \left(
\begin{array}{c}
0 \\
\frac{2 r^2 \left(z^2-2 M^2\right)
r''+r \left(-z^2 \left(z^2-4
M^2\right) \phi '^2-M r'^2 \left(2 M+3
\sqrt{r^2+z^2}\right)+M z'^2
\left(\sqrt{r^2+z^2}-2 M\right)+M
\left(\sqrt{r^2+z^2}-2
M\right)\right)+z \left(z
\left(z^2-4 M^2\right) r''-4 M
\sqrt{r^2+z^2} r'
z'\right)+r^3 \phi '^2 \left(M
\left(2 M+\sqrt{r^2+z^2}\right)-2
z^2\right)+r^4 r''+r^5
\left(-\phi
'^2\right)}{\left(r^2+z^2\right)
\left(-4 M^2+r^2+z^2\right)} \\
\frac{2 r' \phi ' \left(-4 M^2+r^2
\left(1-\frac{2
M}{\sqrt{r^2+z^2}}\right)+z^2\right)
+r \left(\left(r^2-4 M^2\right) \phi
''-\frac{4 M z z' \phi
'}{\sqrt{r^2+z^2}}+z^2 \phi
''\right)}{r \left(-4
M^2+r^2+z^2\right)} \\
\frac{r \left(r \left(r^2-4 M^2\right)
z''-4 M \sqrt{r^2+z^2} r'
z'\right)+2 z^2 \left(r^2-2
M^2\right) z''+M z \left(r'^2
\left(\sqrt{r^2+z^2}-2 M\right)-z'^2
\left(2 M+3
\sqrt{r^2+z^2}\right)+\left(\sqrt{r^
2+z^2}-2 M\right) \left(r^2 \phi
'^2+1\right)\right)+z^4
z''}{\left(r^2+z^2\right) \left(-4
M^2+r^2+z^2\right)} \\
\end{array}
\right) $$
To specifically find a circular orbit we can set $z=0$ and $r=R$ and $\phi = d\phi \ t$. That simplifies the geodesic equation to: $$ 0=\left(
\begin{array}{c}
0 \\
\frac{-\text{d$\phi $}^2 M R^2-\text{d$\phi
$}^2 R^3+M}{2 M R+R^2} \\
0 \\
0 \\
\end{array}
\right) $$ so $$ {d\phi}=\frac{\sqrt{M}}{\sqrt{M
R^2+R^3}} $$
Solving for $\phi = 2\pi$ we get $$ t_{2\pi}=\frac{2 \pi \sqrt{R^2 (M+R)}}{\sqrt{M}} $$
Evaluating proper time along the worldline of the earth we get $$ \frac{{d\tau}}{{dt}}=\frac{\sqrt{-4
M^2-2 M R+R^2}}{\sqrt{M R+R^2}} $$ which we can integrate to get Big Ben's time over one year to get $$\tau_{2\pi} =\int_0^{t_{2\pi}} \frac{d\tau}{dt} dt = 2 \pi R \sqrt{-\frac{4 M}{R}+\frac{R}{M}-2} $$ Plugging in the mass of the sun $M=1480$ in geometrized units (meters) and the orbital radius of the Earth $R=1.496 \ 10^{11}$ we get the proper time $\tau_{2\pi} = 9.45 \ 10^{15}$ which is one year in geometrized units.
OK, now adding the math for the other frame that the OP was going on about. We will boost along the cylindrical axis with the Lorentz transform in that direction so $$t=\frac{T-v Z}{\sqrt{1-v^2}}$$ $$z=\frac{Z-v T}{\sqrt{1-v^2}}$$ Note that the $r$ and $\phi$ coordinates are unchanged. Transforming the line element we get $$ds^2= {d}\phi ^2 \left(r^2-2 r^2
U\right)+{{d}r}^2 (1-2
U)+{{d}T}^2 \left(\frac{2 U
\left(v^2+1\right)}{v^2-1}-1\right)-\frac{8
{dT} {dZ} U
v}{v^2-1}+{dZ}^2
\left(\frac{2 U
\left(v^2+1\right)}{v^2-1}+1\right) $$ And transforming the potential we get $$U=-\frac{M}{\sqrt{r^2 + \left( \frac{Z-vT}{\sqrt{1-v^2}} \right)^2}}$$
With the line element determined in the boosted frame we simply apply the same math as before to obtain the geodesic equations in this frame. We get the following monstrosity (thank you Mathematica) $$0=
\left(
\begin{array}{c}
0 \\
\frac{\left(v^2-1\right)^2 r \phi '^2
\left(-\frac{4 M^2
\left(v^2-1\right)}{(Z-T
v)^2-\left(v^2-1\right)
r^2}-1\right)+\frac{M \left(1-v^2\right)^3
r r'^2 \left(2 M \left(v^2-1\right)+4 v
Z' \sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}-\left(v^2+3\right)
\sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}\right)}{\left((Z-T
v)^2-\left(v^2-1\right)
r^2\right)^2}-\frac{M r' (Z-T v)
\left(-\frac{2 M v \left(\left(v^2+1\right)
Z'^2-4 v
Z'+v^2+1\right)}{\sqrt{r^2-\frac{(Z-
T v)^2}{v^2-1}}}+Z' \left(v
\left(v^2-3\right) Z'+4\right)+v
\left(v^2-3\right)\right)}{\left(r^2-\frac
{(Z-T v)^2}{v^2-1}\right)^{3/2}}+\frac{M
\left(v^2-1\right)^3 r^2 \phi '^2
\left(\sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}-2 M\right) \left(v r'
(Z-T v)+\left(v^2-1\right)
r\right)}{\left((Z-T
v)^2-\left(v^2-1\right)
r^2\right)^2}-\frac{M v
\left(v^2-1\right)^3 r'^3 (T v-Z)
\left(\sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}-2 M\right)}{\left((Z-T
v)^2-\left(v^2-1\right)
r^2\right)^2}+\frac{M \left(1-v^2\right)^3
r \left(\left(v^2+1\right) Z'^2-4 v
Z'+v^2+1\right)
\left(\sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}-2 M\right)}{\left((Z-T
v)^2-\left(v^2-1\right)
r^2\right)^2}}{\left(v^2-1\right)^2
\left(\frac{4 M^2 \left(v^2-1\right)}{(Z-T
v)^2-\left(v^2-1\right)
r^2}+1\right)}+r'' \\
\frac{\phi ' \left(\frac{4 M
\left(v^2-1\right)^2 r r' \left(v
Z'-1\right)}{\left((Z-T
v)^2-\left(v^2-1\right)
r^2\right)^{3/2}}+\frac{M (Z-T v)
\left(\frac{2 M v \left(-\left(v^2-1\right)
r'^2+\left(v^2+1\right) Z'^2-4 v
Z'+v^2+1\right)}{\sqrt{r^2-\frac{(Z-
T v)^2}{v^2-1}}}+v \left(\left(v^2-1\right)
r'^2-v^2+3\right)-v \left(v^2-3\right)
Z'^2-4 Z'\right)}{\sqrt{1-v^2}
\left(r^2-\frac{(Z-T
v)^2}{v^2-1}\right)^{3/2}}\right)-\left(1-v^2
\right)^{3/2} \phi '' \left(-\frac{4 M^2
\left(v^2-1\right)}{(Z-T
v)^2-\left(v^2-1\right)
r^2}-1\right)}{\left(1-v^2\right)^{3/2}
\left(\frac{4 M^2 \left(v^2-1\right)}{(Z-T
v)^2-\left(v^2-1\right)
r^2}+1\right)}+\frac{M v r^2 (T v-Z)
\phi '^3}{\left((Z-T
v)^2-\left(v^2-1\right) r^2\right) \left(2
M+\sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}\right)}+\frac{2 r' \phi
'}{r} \\
\frac{\left(v Z'-1\right) \left(\frac{M
(Z-T v) \left(\frac{2 M
\left(\left(v^2-1\right)
r'^2+\left(v^2-1\right) r^2 \phi
'^2-\left(v^2+1\right) Z'^2+4 v
Z'-v^2-1\right)}{\sqrt{r^2-\frac{(Z-
T v)^2}{v^2-1}}}-\left(v^2-1\right)
r'^2-\left(v^2-1\right) r^2 \phi
'^2+v^2 Z'^2+4 v Z'-3 Z'^2-3
v^2+1\right)}{\sqrt{1-v^2}
\left(r^2-\frac{(Z-T
v)^2}{v^2-1}\right)^{3/2}}+\frac{4 M
\left(v^2-1\right)^2 r r'
\left(v-Z'\right)}{\left((Z-T
v)^2-\left(v^2-1\right)
r^2\right)^{3/2}}\right)}{\left(1-v^2\right)^{3/2} \left(-\frac{4 M^2
\left(v^2-1\right)}{(Z-T
v)^2-\left(v^2-1\right)
r^2}-1\right)}+Z'' \\
\end{array}
\right)
$$
Now, as before we will simplify this substantially by considering only circular orbits which will wind up as helical orbits in this frame. We will use $r=R$ and $\phi = d\phi \ T$ as before, but this time we will have $Z=v T$. These are the transforms of the original circular orbit. With these, the geodesic equation simplifies to $$ 0=\left(
\begin{array}{c}
0 \\
-\frac{M \left(\text{d$\phi $}^2
R^2+v^2-1\right)+\text{d$\phi $}^2 R^3}{R (2
M+R)} \\
0 \\
0 \\
\end{array}
\right) $$
Solving for $d\phi$ we get $$ \text{d$\phi $}=\frac{\sqrt{M-M v^2}}{\sqrt{M
R^2+R^3}}$$ Note that this is slower than $d\phi$ in the other frame by a factor of $1/\gamma=\sqrt{1-v^2}$. So in this frame the angular speed required to maintain a stable orbit (geodesic) is "dilated". This is the key fact that everyone with any experience in GR already knew.
We continue with the rest of the calculations. In this frame the time required to get to $\phi=2\pi$ is $$T_{2\pi}=\frac{2 \pi \sqrt{R^2 (M+R)}}{\sqrt{-M
\left(v^2-1\right)}}$$ which we see is also "dilated". Meaning that the year is longer relative to coordinate time $T$ in the boosted frame. And finally, we plug this back into the line element to find the proper time $$\frac{\text{d$\tau $}}{\text{dT}}=\sqrt{\frac{4
M^2 v^2}{R (M+R)}-\frac{4 M^2}{R
(M+R)}+\frac{2 M v^2}{M+R}-\frac{R
v^2}{M+R}-\frac{2 M}{M+R}+\frac{R}{M+R}}$$ and integrate it over the year to obtain $$\tau_{2\pi}= \int_0^{T_{2\pi}}\frac{d\tau}{dT} dT =2 \pi \sqrt{R \left(\frac{R^2}{M}-4 M-2
R\right)}$$ Which is the exact same expression for proper time as before, and substituting numbers gets the same numbers as before. So Big Ben measures the same amount of time in a year. Both the year and Big Ben are "dilated" the same, and the angular velocity required to maintain a stable orbit matches the actual angular velocity.
So the conclusion was exactly as expected. Big Ben and the year both time dilate the same and the boosted velocity is the correct orbital velocity. Indeed, from first principles it could not be any other way. Any invariants must be the same in all frames. So the conclusion was guaranteed. What was actually tested by the above math was whether or not I can program Mathematica for GR calculations.
EDIT 2:
Based on comments it seems that even after the math of EDIT 1 the OP still has the misunderstanding expressed here:
If the earth does not spiral into the sun then the spacetime curvature in the vicinity of the sun has decreased by the factor 1/γ. From this it follows that spacetime curvature is not invariant, general relativity is wrong and starlight can't bend.
As shown in EDIT 1, the earth does not spiral into the sun in the boosted frame, and the orbit is time-dilated the same as Big Ben. However, the claim that this implies that spacetime curvature has decreased is false. This can be refuted by calculating a complete set of spacetime curvature invariants, such as the Carminati–McLenaghan invariants.
Here I evaluated all of the CM curvature invariants at $R=1.496 \ 10^{11}$ and $M=1480$ corresponding to the orbital radius of the earth and the mass of the sun respectively in geometrized units, as above. The boosted frame was boosted to $v=0.6$ and the curvature was evaluated at $z=Z=T=0$. The curvature is not a function of either $t$ or $\phi$.
$$\begin{array}{ c c c }
\text{Invariant} & \text{Sun's frame} & \text{Ship's frame} \\
\hline
R & 4.37 \ 10^{-38} & 4.37 \ 10^{-38} \\
R_1 & 6.26 \ 10^{-76} & 6.26 \ 10^{-76} \\
R_2 & -7.06 \ 10^{-114} & -7.06 \ 10^{-114} \\
R_3 & 2.05 \ 10^{-151} & 2.05 \ 10^{-151} \\
M_3 & 3.49 \ 10^{-136} & 3.49 \ 10^{-136} \\
M_4 & 1.38 \ 10^{-174} & 1.38 \ 10^{-174} \\
W_1 & 1.17 \ 10^{-60} + 1.17 \ 10^{-60} \ i & 1.17 \ 10^{-60} + 1.17 \ 10^{-60} \ i \\
W_2 & -5.18 \ 10^{-91} -5.18 \ 10^{-91} \ i & -5.18 \ 10^{-91} -5.18 \ 10^{-91} \ i \\
M_1 & -2.37 \ 10^{-106} - 2.37 \ 10^{-106} \ i & -2.37 \ 10^{-106} - 2.37 \ 10^{-106} \ i \\
M_2 & 3.49 \ 10^{-136} + 0 \ i & 3.49 \ 10^{-136} + 0 \ i\\
M_5 & -2.47 \ 10^{-166} + 0 \ i & -2.47 \ 10^{-166} + 0 \ i \\
\end{array}$$
This shows that the curvature is indeed invariant. Despite the changes to the components of the various tensors (including the curvature tensors), none of the invariants are changed (including the curvature invariants). This is a complete set of curvature invariants, so it completely characterizes the curvature tensor at the specified event.
That it must come out this way is a foregone conclusion from first principles, but it is good to see it demonstrated.
There are two additional facts that are worth noting with respect to this specific problem. First, the fact that all of these curvature invariants are very small supports the validity of the weak field approximation in the sun's frame. Second, the fact that they are the same in the ship's boosted frame shows that the boost is not an additional approximation. So, the weak field metric is an approximation, but it is a valid one in this problem, and the boost is not an additional approximation.
