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Thread: Equations for ballistic travel

  1. #1 Equations for ballistic travel 
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    May 2005
    I am trying to work out a generic equation for ballistic travel on a sphere (ie, the earth) from a given initial velocity in the forward and upward direction. How does one calculate distance traveled along the surface of the sphere for a given initial velocity in the X and Y direction? Any ideas? It seems like it should be relatively easy to work out, but unfortunately my calculus is very rusty...

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  3. #2  
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    Apr 2007
    It can get pretty complex depending how accurate it needs to be and whether it is relatively short range or long range (where Coriolis effect and such have to be considered).

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  4. #3  
    Forum Radioactive Isotope mitchellmckain's Avatar
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    Oct 2005
    Salt Lake City, UTAH, USA
    Well for one thing, it would not be appropriate to treat gravitational acceleration as a constant on a scale where the curvature of the earth is significant. So for this you should start with universal law of gravitation

    F = G M m / r^2 (where G is the universal gravitational constant)


    F = - G M m r / r^3 using vectors

    This looks simpler than it really is, because in spherical coordinates the acceleration in Newton's second law F=ma is given by.

    a = d^2(r)/dt^2 = (d^2 r/dt^2 - r (d theta / dt)^2) r/r + (r d^2 theta / dt^2 + 2 dr/dt d(theta)/dt) theta/theta

    and thus the differential equations to solve are

    r^2(d^2 r/dt^2 - r (d theta / dt)^2) = -GM


    r d^2(theta)/dt^2 + 2 dr/dt d(theta)/dt = 0

    The solution is basically Celestial mechanics and the solution to this will be the same as Keplers laws of planetary motion. Assuming you do not have an orbit you are not interested in those concerning period, but only position as a function of time. In any case, a look at Wikipedia on Kepler's laws will show you that this is probably not as trivial as you thought. Trajectories will be elliptical orbits interrupted by the surface of the earth.

    One of the famous results was that r^2 d(theta)/dt is a constant of motion which is what makes the area swept out by the radius equal in equal times.

    Anyway, perhaps you can see why approximations are used whenever possible.
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  5. #4  
    Moderator Moderator Janus's Avatar
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    Jun 2007
    Here's one way to work it out.

    It can be done in steps.

    First find the semi major axis of the elliptical trajectory the projectile will take with:

    a = -GM/(v²-2GM/r)

    v is the velocity of the projectile
    r is the radius of your sphere
    M is the mass of the sphere
    With this you can find the eccentricity of the trajectory with:

    e = sqrt(1-r²(v_h)²/(aGM))

    where v_h is the horizontal velocity of the projectile.

    Now you find Phi, the angle from periapis at which the trajectory intersects with the surface of the sphere:

    theta = acos[{a(1-e²)/r-1}/e]

    180-theta gives us the angle to apoapis (the top of the trajectory).


    d= pi*r(1-theta/180)

    Will give us our distance.

    Projectile speed of 1000 m/s fired at an angle of 45° from the Surface of the Earth.

    a= -6.673e-11(5.97e24)/(1000²-2(6.673e-11)(5.97e24)/6.378e6 )= 3.215e6 m

    e = sqrt(1-(6.378e24)²(707.1)²/(3.215e6(6.673e-11)(5.97e24)))= 0.992

    theta = acos[{3.215e6(1-0.992²)/6.378e6-1}/0.992] = 179.5°


    d= 2pi*6.378e6(1-179.5/180) =111317m = 111.3km

    This does not take the rotation of the Earth and its effect of the trajectory into account.
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  6. #5  
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    May 2005
    A belated thanks to mitchellmckain and Janus for the helpful replies. I asked about this because I recently became interested in the concept of "sub-orbital" space planes that could carry passengers across long distances in very short times by making hypersonic hops out of the atmosphere. If I'm doing my math right, it looks like you would need about 6 km/sec to get from England to New York on a ballistic flight. If your engine has a specific impulse of around 420 seconds, that would mean your "space plane" only needs a mass ratio of around 4 - which seems very doable. The flight would take something like 20-30 minutes, depending on how long the initial takeoff and landing lasted.

    Of course, the ticket price might be pretty high. I wonder how many people would want to ride in such a thing simply for the thrill?
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