$$ds^2 = -\left(1 - \frac{2GM}{r}\right) dt^2 + \left(1 - \frac{2GM}{r}\right)^{-1} dr^2 + r^2 d\Omega^2$$
$$\frac{d^2x^\mu}{d\lambda^2} + \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\lambda} \frac{dx^\beta}{d\lambda} = 0$$ moore general relativity workbook solutions
After some calculations, we find that the geodesic equation becomes $$ds^2 = -\left(1 - \frac{2GM}{r}\right) dt^2 + \left(1
$$\frac{d^2r}{d\lambda^2} = -\frac{GM}{r^2} \left(1 - \frac{2GM}{r}\right) \left(\frac{dt}{d\lambda}\right)^2 + \frac{GM}{r^2} \left(1 - \frac{2GM}{r}\right)^{-1} \left(\frac{dr}{d\lambda}\right)^2$$ moore general relativity workbook solutions
which describes a straight line in flat spacetime.
The gravitational time dilation factor is given by