How close do you have to come to the Earth to be influenced by its gravity? (Advanced)

How close would an object, travelling at a speed V far from the Earth (Mass M) come if it were influenced by Earth's gravitational field ?

Thank you for your question; it is a very interesting problem.

There are several way to answer your question, the simplest is to use what is called an "hyperbolic fly-by" assumption. Let us assume that the object is already so close to the Earth that the terrestrial gravitational influence dominates over the solar. The asteroid is coming with a velocity with respect to the Earth called velocity at infinity vinfinity, and, since the object is not bounded, i.e., has positive energy, its trajectory is going to be a hyperbola, with the focus in the center of mass of the Earth (it would have been an ellipse if the object were bounded, i.e., had negative energy, like an artificial satellite, or a parabola if the object had just enough energy to escape the gravitational attraction of the Earth. Ellipse, circle, parabola, and hyperbola are called conic sections, since can be obtained from "cutting" a cone with a plane). Any orbit, elliptical, parabolic, or hyperbolic, requires six parameters to be described in space (essentially you have an object in a tridimensional space, so you need six constants, three for the positions and three for the velocities). For our purposes, we just need to introduce two of them, the eccentricity and the semimajor axis. The eccentricity is a parameter that tells you how "open" is a conic section: is zero for a circle, < 1 for an ellipse, > 1 for a hyperbola. The bigger the eccentricity, the more "open" the hyperbola. The semimajor axis tells you how "large" the orbit is. For a circle it would correspond to the radius and for an ellipse it would be the length of the longer axis.

For a hyperbola it is more difficult to visualize, since the hyperbola is a open trajectory. It is mathematically defined such that any point on a hyperbola has the difference of the distances from two fixed points called focii equal to 2*a (for an ellipse it would have been the sum of the distances equal to that value).

If we call the minimum distance q, it is possible to show that for any conical section.

(1) q=a(1-e)

Since e > 1 and q has to be positive (it's a distance...) that's the reason why for an hyperbola a is usually considered negative in celestial mechanics. What you are looking for is the value of q, but to compute that we first need to know what are the expressions of a and e in terms of what we know, the velocity at infinity vinfinity, the impact parameter (see figure) and the mass of the Earth.

Under the (big) assumption that the perturbations from the Sun are negligible, the motion is planar and the angular momentum, i.e., the vector product of the particle position time its velocity, is conserved. So, its initial value would be a constant during all the flyby.

If we called D what you called the impact parameter, then the angular momentum h at t=0 would be given by:

(2) h=vinfinityD

and is going to point along the z-axis. It is possible to show that, for any conic orbit (i.e., an ellipse, an hyperbola, or a parabola, see Danby 1998, Fundamentals of celestial mechanics, Ed. Willmann-Bell, pp. 78-82 for example) h is given by:

(3) h=sqrt{μ*a*(1-e2)}

where μ is the product of the gravitational constant times the mass m2 of the Earth. Equating (2) with (3) gives you a relation between m2, vinfinity, D, which are known, with a and e which are unknown, i.e, (4) sqrt{μ*a*(1-e2)}=vinfinityD

We need another equation, and that's given by the conservation of energy. If we consider an asteroid of mass 1, at any point its energy would be given by:

(5) E=-μ/r+v2/2

i.e., the usual sum sum of the potential and kinetic energy. At t=0 v=vinfinity and r, distance from the center, is so big that the first term can be neglected. For any orbit, its energy can also be defined as (Danby, pp. 63-65):

(6) E=-μ/2a

We are finally done: equating (5) and (4) (at infinity) gives:

(7) a=-μ/vinfinity2

(which is negative, as it should be for a hyperbola). Putting (7) into (4) yields, after some algebra:

(8) e2=1+vinfinity^4*D22

So, q is finally given by:

(9) q=-Gm2/vinfinity(1-sqrt{1+vinfinity4*D22}

This is the expression you were looking for, which is given in terms of the mass of the Earth, the velocity at infinity, and the impact parameter.

In deriving this expression, however, we have neglected solar perturbations. If we wanted to be more realistic, we should have also considered the influence of the Sun. This would have been what in celestial mechanics we call the "three-body" problem, for which, unfortunately, there are not exact analytical solutions (but approximated ones can be found by using several mathematical tecniques). One way to do that could have been to use symplectic integrators, that are programs that numerically compute the evolution of the orbital elements of objects subjected to the gravitational influences of one central body and several perturbers.

Your problem is however very difficult, since the asteroid is passing from the gravitational influence of the Sun to a region in which Earth gravity prevails. It was just very recently that new symplectic integrators have been written to address this problem (SyMBA by Levison and Duncan, Mercury by J. Chambers), and this has permitted to investigate several interesting problems, like for example the long term (~500 Myr) effects of close encounters between large asteroids and members of asteroid families.

This page was last updated on July 18, 2015.

About the Author

Valerio Carruba

Valerio is currently a Professor at the São Paulo State University in Brazil (UNESP), and he mostly works with asteroid dynamics.  He went to college in Italy at the University "La Sapienza", took his Ph. D. in Qstronomy at Cornell University, and then went to Brazil in 2004 for various pos-docs that then "evolved" into his current permanent position.

More information on Valerio and his research is available at:

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