The escape speed from a planet of mass $M$ and radius $R$ is $v_1$. If the planet's mass becomes $4M$ while radius remains $R$, the new escape speed $v_2$ is:

Newton's law of gravitation: $F = G\frac{m_1 m_2}{r^2}$ where $G = 6.674\times10^{-11}$ N m$^2$ kg$^{-2}$. Gravitational field: $g = GM/r^2$ (at distance $r$). Surface gravity: $g_0 = GM/R^2$. Gravitational PE: $U = -GMm/r$ (negative, attractive). At infinity: $U = 0$. At surface: $U = -GMm/R$.

Circular orbit at distance $r$: centripetal force = gravitational force: $mv_o^2/r = GMm/r^2$ → $v_o = \sqrt{GM/r}$. Time period: $T = 2\pi r/v_o = 2\pi\sqrt{r^3/GM}$. Kepler\'s third law: $T^2 \propto r^3$. Geostationary orbit: $T = 24$ hr, $r \approx 42,000$ km from Earth centre, always above same point on equator.

$U(r) = -GMm/r$. At surface: $U = -GMm/R$. Binding energy = energy needed to remove from surface to infinity = $+GMm/R = \frac{1}{2}mv_{esc}^2$. Potential $V = -GM/r$ (field per unit mass analogy). $g = -dV/dr$. For a hollow sphere: $V = -GM/R$ inside (constant), field = 0 inside.

Kepler's three laws of planetary motion: First (Ellipse law): planets move in elliptical orbits with sun at one focus. Second (Area law): line joining planet to sun sweeps equal areas in equal times (conservation of angular momentum). Third (Period law): $T^2 \propto a^3$ where $a$ = semi-major axis. For circular orbit: $T^2 = \frac{4\pi^2}{GM}r^3$. All three are consequences of Newton's law of gravitation.

Above the surface, $g_h = g\left(1 - \frac{2h}{R}\right)$ for $h \ll R$; below it, $g_d = g\left(1 - \frac{d}{R}\right)$. Note the factor of 2 appears only in the height case, so g falls twice as fast going up as going down. At the centre of the Earth g is zero, since the shell of mass outside a point exerts no net gravitational force on it. Rotation adds a further variation with latitude, $g_\lambda = g - \omega^2 R\cos^2\lambda$, which makes g greatest at the poles and least at the equator.

The general expression is $U = -\frac{GMm}{r}$, negative because the zero of potential energy is taken at infinite separation and the force is attractive — work must be done on the system to pull the bodies apart. The familiar $mgh$ is only the approximation valid near the surface, where the change in r is small compared with R. A satellite in a bound orbit therefore always has negative total energy, $E = -\frac{GMm}{2r}$, and reaching zero total energy is precisely what escaping means.

An astronaut in orbit is not beyond gravity — at the height of the space station, g is still about 89% of its surface value. The sensation of weightlessness arises because the astronaut and the station are both in free fall towards the Earth at the same rate, so there is no normal reaction between them. Apparent weight is the normal reaction, not the gravitational force, which is why a lift accelerating downward at a makes a passenger feel lighter by $ma$, and one in free fall makes them feel nothing at all. True weightlessness would require being infinitely far from every mass, which is never the case.