Search results
Results from the WOW.Com Content Network
Neither orbit is perfectly circular; Earth has an orbital eccentricity of 0.0168, and Mars of 0.0934. The two orbits are not quite coplanar either, as the orbit of Mars is inclined by 1.85 degrees to that of Earth. The effect of the gravity of Mars on the cycler orbits is almost negligible, but that of the far more massive Earth needs to be ...
All bounded orbits where the gravity of a central body dominates are elliptical in nature. A special case of this is the circular orbit, which is an ellipse of zero eccentricity. The formula for the velocity of a body in a circular orbit at distance r from the center of gravity of mass M can be derived as follows:
A lunar cycler or Earth–Moon cycler is a cycler orbit, or spacecraft therein, which periodically passes close by the Earth and the Moon, using gravity assists and occasional propellant-powered corrections to maintain its trajectories between the two. If the fuel required to reach a particular cycler orbit from both the Earth and the Moon is ...
The Oberth effect is used in a powered flyby or Oberth maneuver where the application of an impulse, typically from the use of a rocket engine, close to a gravitational body (where the gravity potential is low, and the speed is high) can give much more change in kinetic energy and final speed (i.e. higher specific energy) than the same impulse ...
It takes 250 days (0.68 years) in the transit to Mars, and in the case of a free-return style abort without the use of propulsion at Mars, 1.5 years to get back to Earth, at a total delta-v requirement of 3.34 km/s. Zubrin advocates a slightly faster transfer, that takes only 180 days to Mars, but 2 years back to Earth in case of an abort.
The two-body problem in general relativity (or relativistic two-body problem) is the determination of the motion and gravitational field of two bodies as described by the field equations of general relativity. Solving the Kepler problem is essential to calculate the bending of light by gravity and the motion of a planet orbiting its sun.
When the vehicle has left the gravity well, it is traveling at a speed V = Δ v 1 + 2 V esc Δ v . {\displaystyle V=\Delta v{\sqrt {1+{\frac {2V_{\text{esc}}}{\Delta v}}}}.} For the case where the added impulse Δ v is small compared to escape velocity, the 1 can be ignored, and the effective Δ v of the impulsive burn can be seen to be ...
But in those cases, as with elliptical orbits, the area swept out by a chord between the attractor and the object following the trajectory increases linearly with time. For the hyperbolic case, there is a formula similar to the above giving the elapsed time as a function of the angle (the true anomaly in the elliptic case), as explained in the ...