Blog #24 - Hill Spheres

Hill Spheres

One outcome of planet formation is systems of satellites around planets. Now you may ask yourself, why do some planets have moons 10s of millions of kilometers away, while the Earth’s moon is only 400,000 km away. To answer this question we need to think about how big of a region around a planet is dominated by the gravity of a planet, i.e. the region where the gravitational pull of the planet is more important than the gravitational pull of the central star (or another planet). 

(A) Gravitational forces. Put a test mass somewhere between a star of mass Ms and a planet of mass mp at a distance rp from the star. Make a drawing marking clearly these characteristics as well as the distance r between the test particle and the planet. Write separate expressions for the gravitational force on the particle from the star and on the particle from the planet. At what distance r from the planet are the two forces balanced? This distance approximates the radius of the Hill sphere, which in the case of planet formation is the sphere of disk material which a planet can accrete from. 

We begin by drawing out the situation described in the problem 

From the drawing we need to balance gravitational forces between the star and the planet both acting on the orbiting particle or satellite. It's important to note that the balance has to exist on all sides of the particle's orbit around the planet. \[\frac{G M_s}{(R_p - R)^2} - \frac{G M_s}{(R_p + R)^2} = \frac{G M_p}{R^2} \] Cancelling G and simplifying both sides we get \[\frac{4 M_s R_p R}{R_p^4 - 2 R_p^2 R^2 + R^4} = \frac{M_p}{R^2} \] Because R is much smaller than \(R_p\) we can simplify the left side of the equation to \[\frac{4 M_s R}{R^3} = \frac{M_p}{R^2} \] Then by rearranging to solve for R we get \[R = R_p (\frac{M_p}{4 M_s})^{\frac{1}{3}}\]However the true equation for Hill Sphere differs by a constant factor due to orbital rotation speeds and comes out to \[R = R_p (\frac{M_p}{3 M_s})^{\frac{1}{3}}\]

Plugging constants from the table into the equation above, we get \[R_{Earth} = 1.49*10^6 \space km \quad R_{Jupiter} = 5*10^7 \space km \quad R_{Neptune} = 1.15*10^8 \space km \]Comparing these values to the values for the furthest moon around each planet it becomes clear that each moon is well within the planet's Hill Sphere.