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Gravity Assist Oberth Effect Sun

  1. Sep 28, 2012 #1


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    Hi all,

    I'm having some trouble understanding conflicting information on the web.

    1) I know that a gravitational assist on a spaceship around a planet can increase the spaceship's speed with respect to the solar system. A gravitational flyby will not work for ships that stay within the solar system if the sun is used, because the sun is at rest with respect to the solar system.

    2) However, there is another effect, the Oberth Effect, where, if one fires the ship's engines at point of closest aproach to the planet assiting the flyby, this different effect will also boost the ship's speed. It seems that such an Oberth Effect CAN boost a ship's speed as it passes around the sun, and on one TV show I saw, up to 300 km/sec. It is not used because no technology can handle the heat of the sun were a ship to fly close enough.

    Is the statement 2) wrong? If it is right, does anyone know how to calculate what the boost in speed a ship would have in an Oberth Effect around the sun, or any reliable websites that provide accurate numbers or formula? Is the Oberth Effect dependant on the Gravity Assist in anyway, or are they really two, distinct phenomenon? (assume our ship has the heat shield)
  2. jcsd
  3. Sep 28, 2012 #2


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    Statement 2 is correct, and comes from the fact that for the same delta v, the energy change is much higher if the starting velocity is higher. Rocket engines with a given amount of fuel provide a fixed change in velocity, rather than a fixed change in energy, so this can be used to the ship's advantage by gaining as much velocity as possible through passive (gravitational) means before firing the ship's engines.
  4. Sep 30, 2012 #3


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    Interesting. Would I therefore be correct in saying that: when a spaceship goes around a planet, it can increase its energy tremendously from two very powerful effects: 1) a gravity assist AND 2) an Oberth Effect since they are both seperate effects? Also, a ship flyby around the sun would only benefit from the Oberth Effect.
  5. Oct 1, 2012 #4


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    I've read all about the two above effects now and understand they are seperate phenomenon. There's still one thing I can't do quite yet. I think the answer is easy, but I can't see it.

    Suppose my ship approaches a planet and wants to take advantage of both effects. It wants that 2U+v pickup in speed from the gravitational assist, AND he wants to fire his engines to gain that delta-v pickup from Oberth as well. How do I calculate the final new speed from using both effects?

    I do now, finally, know of websites now that discuss the gain in velocity of one effect or the other, but not both at the same time. Seems to me that anyone would naturally want to utilize both effects given how much fuel is needed to get anywhere in the solar system quickly, and how much each method alone can contribute so much to a spaceship's energy for free, so I'm surprised other people haven't posted about combining the two effects on the web.
  6. Oct 18, 2012 #5


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    Hi all,
    After weeks of diligent searching, I finally found an answer to what the Oberth Effect is. It turned out to be pretty simple. I don’t need any help with what I am posting, rather, I felt I should put my answer to my question to share with the community. It was so tough to find out about, hopefully this post will make searching easier for the next peron. Okay, here it is, ballpark calc of Oberth Effect:

    A space ship at r = ∞ is heading towards a planet on a hyperbolic trajectory. At r = ∞, his U is zero, and his kinetic energy is (1/2)msVi^2. “Vi” is the ship’s velocity at r = ∞. That is, all of his energy is kinetic. Now, we want to burn at periapsis—as close as possible to the planet. At the periapsis, we will have both gravity potential energy and kinetic energy, and we will be travelling at v_p (velocity at periapsis) We conserve energy between the positions r=∞ and r=r_p (“r” at periapsis or the closest approach to the planet or sun). At r = ∞, the gravitational potential energy is zero and at periapsis the gravitational potential energy is -GMm/r_p. We write:

    (at r = ∞) (at r=r_p)
    (½)m(Vi)^2 = (½)m(V_p)^2-GMm/r_p

    Where G is the gravitational constant, and M is the mass of the body we do this powered slingshot off of. The mass of the spaceship “m” does not matter, and cancels out and we multiply everything by 2. We get:
    (Vi)^2 = (V_p)^2-2GM/r_p

    Now, “2GM/r_p” is the square of the escape speed of that particular periapsis for a particular planet. So we re-write the above as:

    (Vi)^2 = (V_p)^2-V_esc^2

    Notice that we square these velocities and aren’t adding them. Why? Because the above is from conservation of energy, not addition of velocities and the v_esc is more of an energy term. Next step, solve for the velocity of the spaceship at periapsis, just before you burn.

    We have

    SQRT((Vi)^2+-V_esc^2) = V_p

    Now, I burn with delV, and my new velocity just after periapsis is V_p + delV and we modify the above for our velocity at periapsis just after the burn.

    SQRT((Vi)^2+-(V_esc)^2)+delV = V_p’

    Now, the goal: what is your new speed at r = ∞? Travelling towards the planet, the ship speeds up until it hits periapsis, and travelling away from the planet, it would slow down again to its original speed it had before encountering the planet, but this burn converted gravitational potential energy of the gas into kinetic energy. At periapsis, both the mass of the ship and mass of the gas have had their potential energy converted to kinetic energy. This is NOT the chemical energy of the gas. The gas has mass, so it had potential energy when the ship brought it with it to periapsis, and that gravitational potential energy is now kinetic. The ship gives the gas a shove out of the ship at periapsis, and the gas reacts back on the ship via Newton’s 3rd Law (“Rocket Equation” explains this part). The ship accelerates at periapsis and gains kinetic energy. The gas stays at periapsis, having surrendered its gravitational potential energy to the ship. Thus, to find the new speed of the ship at r = ∞, we conserve energy again between periapsis with our new kinetic energy after the burn and our exit r = ∞. We write:

    (at r=r_p) (at r = ∞)

    (½)m(V_p+delV)^2-GMm/r_p = (½)m(V_∞)^2

    Algebra, turning the potential energy term at periapsis into V_esc^2, and rearranging we have:

    (V_p+delV)^2- V_esc^2 = (V_∞)^2.

    “V_p” above is the V_p before the burn, so we plug in that expression here to get:

    sqrt((Vi)^2+(V_esc)^2) +delV)^2- V_esc^2 = (V_∞)^2

    Taking the square root again will get you what your new speed at r = ∞.

    Here’s an example calculation: If a ship at r = ∞ with Vi= 3.2 km/sec approaches to within 200,000 km of the sun (periapsis—closest approach) and burns with 6 km/sec at periapsis, its final speed some distance away after slingshotting from the sun will be 50 km/sec, which is a huge gain compared to the 3.2 km/sec that the ship started with. Ref: Atomic Rockets Webpage. I have other references, but I asked them by email and not sure if they want to be contacted about this.

    Okay, so that’s the Oberth Effect.
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