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the claim
The Earth follows an elliptical trajectory because its orbital velocity is less than the escape velocity.
the verdict
SUPPORTED
the evidence backs this
refutedsupported
the weight of evidence
3 sources for · 0 against
AS REPORTEDno primary record reached; this is what the reporting says

Reference materials confirm that objects following elliptical orbits maintain a speed less than the escape speed at their current distance from the primary body.

Evidence for · 3
cited by 0
the gravitational influence of the primary. If an object is in a circular or elliptical orbit, its speed is always less than the escape speed at its current In celestial mechanics, escape velocity or escape speed is the minimum speed needed for an object to escape from contact with or orbit of a primary body, assuming: Ballistic trajectory – no other forces are acting on the object, such as propulsion and friction No other gravity-producing objects exist. Although the term escape velocity is common, it is more accurately described as a speed than as Ballistic trajectory – no other forces are acting on the object, such as propulsion and friction No other gravity-producing objects exist. Although the term escape velocity is common, it is more accurately described as a speed than as a velocity because it is independent of direction. Because gravitational force between two objects depends on their combined mass, the escape speed also depends on mass. For artificial satellites and small natural objects, the mass of the object makes a negligible contribution to the combined mass, and so is often ignored. Escape speed varies with distance from the center of the primary body, as does the velocity of an object traveling under the gravitational influence of the primary. If an object is in a circular or elliptical orbit, its speed is always less than the escape speed at its current distance. In contrast if it is on a hyperbolic trajectory its speed will always be higher than the escape speed at its current distance. (It will slow down as it gets to greater distance, but does so asymptotically approaching a positive speed.) An object on a parabolic trajectory will always be traveling exactly the escape speed at its current distance. It has precisely balanced positive kinetic energy and negative gravitational potential energy; it will always be slowing down, asymptotically approaching zero speed, but never quite stop. Escape velocity calculations are typically used to determine whether an object will remain in the gravitational sphere of influence of a given body. For example, in solar system exploration it is useful to know whether a probe will continue to orbit the Earth or escape to a heliocentric orbit. It is also useful to know how much a probe will need to slow down in order to be gravitationally captured by its destination body. Rockets do not have to reach escape velocity in a single maneuver, and objects can also use a gravity assist to siphon kinetic energy away from large bodies. Precise trajectory calculations require taking into account small forces like atmospheric drag, radiation pressure, and solar wind. A rocket under continuous or intermittent thrust (or an object climbing a space elevator) can…
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The analysis

rails:sufficiency:supported:for=2+1p:against=0+0p | v55:sufficiency

More for · 2
2016 · cited by 0
why it moves in a elliptical orbit when it is moving with velocity more than the orbital velocity The escape speed $v_{escape}$ is the minimum speed needed, for the object satellite to not follow an orbit. So, if the speed is higher than circular orbital speed (so it will not be a circle) but lower than escape velocity (so it will follow an orbit), then it will have to follow another kind of orbit, which happens to be elliptical. source: http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/orbits.html why is there a specific value of escape velocity Shoot something away extremely fast and it will fly onwards away from Earth and never come back. If it was shot off fast enough, it might get so far away that it doesn't really feel gravity anymore of any significant amount. Now instead throw something upwards. It will fall back. It has clearly not escaped gravity. Somewhere in between the
2022 · cited by 0
Sadly, the teacher made a mistake. In the teachers equation, the kinetic energy is $mv^2/2$ and the second term should be potential energy, that is, without the 2 in the denominator. The escape velocity at any distance from the center of the earth is $\sqrt2$ times the circular orbit velocity. The direction for escape does not matter (unless the object runs into the planet), but the so-called orbital velocity needs to be perpendicular to the radius from the center. Different angles and velocities give elliptical orbits.
Everything we examined (3) — 2 independent sources
This check searched the claim as stated. It did not run a separate search for evidence against it.
  1. Escape velocityreferenceno side taken
  2. Why do the satellites revolves on a circular path around a planet at orbital velocity?referencesame source L2no side taken
  3. Escape Velocity from orbitreferencesame source L2no side taken
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