- #1

kaminasimon

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(0.5 ×mv^2) +(0.5×mV^2)-(GMm÷(r^2))>=0

Which v,V are linear velocity and velocity of thing we launched respectively

We obtain,...

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- Thread starter kaminasimon
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In summary: Thanks for answerNo, the escape velocity is not measure with respect to the moving ground. The escape velocity is defined relative to the center of mass reference frame and is therefore independent of the direction.

- #1

kaminasimon

- 7

- 0

(0.5 ×mv^2) +(0.5×mV^2)-(GMm÷(r^2))>=0

Which v,V are linear velocity and velocity of thing we launched respectively

We obtain,...

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- #2

kaminasimon

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V^2>=(GM÷R)-[(4π^2)÷T^2]×R^2 and result is Min V= 7.893961 km/s

- #3

dauto

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- #4

Dale

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- #5

kaminasimon

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- #6

kaminasimon

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Thanks for answer

- #7

Bandersnatch

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But it doesn't. The escape velocity, despite its name, is actually escapekaminasimon said:I know.The escape velocity depends on the angle we launched. I forgot tell you i assumming the angle is 90 (perpendicular to surface)

EV value is dependent only on the mass producing the gravitational field and the distance from that mass.

What DaleSpam is saying, is merely that once you've calculated the value of the EV, you can

- #8

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No, it doesn't. Escape velocity is a bit of a misnomer; a better name would be escape speed. Escape velocity is not a vector. It is a scalar function of distance from the center of the gravitating body (the Earth in this case), the mass of the gravitating body,kaminasimon said:I know. The escape velocity depends on the angle we launched.

An object launched with a velocity whose magnitude is exactly equal to the escape velocity will just barely escape the gravitational clutches of the gravitating body. Anything less than escape velocity means the object will remain gravitationally bound to the gravitating body. Anything more than escape velocity means velocity remain non-zero as the distance tends to infinity.

- #9

kaminasimon

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Thank you very much

- #10

kaminasimon

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I think it depends on the angle and absolutely the mass escapes the gravitational field of earth. If the angle is enough small to escape the field the the velocity we supplied will be smaller. Because Vector(V)= vector(v1)+vector(v2). The module V^2=(v1)^2+(v2)^2+2(v1)(v2)cos(c) where *c* is angle form vector(v1) and vector(v2). V1 is velocity we supplied and v2 is linear velocity. So v1 will be smaller than. It is depends reference frame we chosen. In my reference frame i chosen is station reference in the universe.

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- #11

Dale

Mentor

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This is not correct. The escape velocity does not depend on the angle, only the speed.kaminasimon said:I think it depends on the angle

- #12

dauto

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kaminasimon said:I think it depends on the angle and absolutely the mass escapes the gravitational field of earth. If the angle is enough small to escape the field the the velocity we supplied will be smaller. Because Vector(V)= vector(v1)+vector(v2). The module V^2=(v1)^2+(v2)^2+2(v1)(v2)cos© which© is angle form vector(v1) and vector(v2). V1 is velocity we supplied and v2 is linear velocity. So v1 will be smaller than. It is depends reference frame we chosen. In my reference frame i chosen is station reference in the universe

You're getting confused because you think the scape velocity is measure with respect to the moving ground in which case it depends on the direction as you said. But the escape velocity is actually defined relative to the center of mass reference frame and is therefore independent of the direction.

- #13

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Note that your post contains ©, where you obviously meant (c). I find it to be rather hard to convince the autocorrect that, yes, I really did mean "c between two parentheses" rather than the copyright symbol. I fixed your post.Regarding the technical details of your post, escape velocity is just a speed. It is independent of direction, independent of current velocity. That said, if an object is already moving at some velocity, achieving escape velocity is most easily accomplished by making the change in velocity, the Δv, be parallel to that initial velocity, and is hardest if the change in velocity is directed against the initial velocity.kaminasimon said:I think it depends on the angle and absolutely the mass escapes the gravitational field of earth. If the angle is enough small to escape the field the the velocity we supplied will be smaller. Because Vector(V)= vector(v1)+vector(v2). The module V^2=(v1)^2+(v2)^2+2(v1)(v2)cos© which© is angle form vector(v1) and vector(v2). V1 is velocity we supplied and v2 is linear velocity. So v1 will be smaller than. It is depends reference frame we chosen. In my reference frame i chosen is station reference in the universe

- #14

kaminasimon

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The law of conservation of energy states that energy cannot be created or destroyed, but can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant.

Escape velocity is the minimum velocity an object must have to escape the gravitational pull of a larger body. According to the law of conservation of energy, the energy of the object must be equal to or greater than the potential energy it has due to the gravitational pull of the larger body. Therefore, conservation of energy is essential in calculating escape velocity.

The two forms of energy involved in calculating escape velocity are kinetic energy and potential energy. Kinetic energy is the energy an object possesses due to its motion, and potential energy is the energy an object has due to its position or state.

The conservation of energy equation, which states that the total energy of a system remains constant, can be applied to calculate escape velocity. The equation takes into account the kinetic and potential energy of the object at the surface and at escape velocity, and sets them equal to each other to solve for the escape velocity.

Escape velocity is crucial in space travel and exploration because it determines the minimum velocity required for a spacecraft to break free from the gravitational pull of a celestial body. Without enough escape velocity, a spacecraft cannot reach its destination or enter orbit around a planet or moon. Thus, understanding and calculating escape velocity is essential for successful space missions.

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