Does the angle at which an object is fired affect its escape velocity?

In summary: The escape velocity of an object on the surface of the Earth is the minimum speed required for it to "break free" from the Earth's gravitational field without requiring further propulsion. Its value is mathematically determined to be v = \sqrt{\frac{2GM}{R}} with M being the mass of the Earth and R being the Earth's radius. If the object is fired pointing radially away from the Earth, then it will travel in a straight line and certainly reach "infinity". If the object is fired pointing radially towards the Earth, it will travel in a straight line and simply accelerate towards the Earth.
  • #1
Bipolarity
776
2
The escape velocity of an object on the surface of the Earth is the minimum speed required for it to "break free" from the Earth's gravitational field without requiring further propulsion.

Its value is mathematically determined to be [itex] v = \sqrt{\frac{2GM}{R}} [/itex] with M being the mass of the Earth and R being the Earth's radius.

If the object is fired pointing radially away from the Earth, then it will travel in a straight line and certainly reach "infinity".

If the object is fired pointing radially towards the Earth, it will travel in a straight line and simply accelerate towards the Earth.

My question is what happens if the object is fired at some angle (with the escape speed of course). Most sources say that if it is fired with some upward angle, it will still reach out to "infinity" because of energy considerations, and the fact that energy is scalar.

But if the object is fired with some downward angle, it will not reach out to "infinity" and simply spiral towards the Earth.

What if the object is fired perpendicular to the radial line?

And how can we prove these results using a rigorous approach, rather than a merely observation that energy is a scalar? I am more of a mathematician, so would prefer a very rigorous attempt to address this question.

All help is appreciated. Thanks!

BiP
 
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  • #2
Hi Bipolarity! :smile:
Bipolarity said:
If the object is fired pointing radially towards the Earth, it will travel in a straight line and simply accelerate towards the Earth.

But if the object is fired with some downward angle, it will not reach out to "infinity" and simply spiral towards the Earth.

No, it will follow a parabola.

If the parabola intersects the earth, of course the object will stop!

If the parabola doesn't intersect the earth, the object will "reach infinity" exactly as if it had been fired outward.
And how can we prove these results using a rigorous approach, rather than a merely observation that energy is a scalar?

That is rigorous. :redface:

Energy shows perfectly rigorously that an object launched at escape velocity will "reach infinity at zero speed", no matter what the angle of launch.
 
  • #3
tiny-tim said:
Hi Bipolarity! :smile:No, it will follow a parabola.

If the parabola intersects the earth, of course the object will stop!

If the parabola doesn't intersect the earth, the object will "reach infinity" exactly as if it had been fired outward.That is rigorous. :redface:

Energy shows perfectly rigorously that an object launched at escape velocity will "reach infinity at zero speed", no matter what the angle of launch.

Thanks tim, but I don't understand how these claims are rigorous. How can you prove them? Suppose I wanted to prove it without using energy...

BiP
 
  • #4
As you get further from the earth, you get closer and closer to flying directly away from it regardless of what angle you started from.
 
  • #5
Energy Proof (simple)

This is basically the proof that's given on Wikipedia, very simple and very easy. I' can't prove the parabola thing that tiny-tim wrote about, so you'll have to figure that one out some other way. However, since energy is scalar, as long as you don't run into the planet (like tiny tim said) the energy is conserved as shown.

We start with conservation of energy.

[tex] (KE + PE)_1 = (KE + PE)_2 [/tex]

Initial kinetic energy is some velocity, we'll call it ve for escape velocity. Likewise, its potential energy is whatever it is on the surface, at radius r from the planet center. As we go to infinity, the potential energy goes to zero as r goes to infinity, and we'll pretend the kinetic energy goes to zero at the point where r goes to infinity.

Additional notes: m is mass of escaping object, M mass of planet, G gravitational constant.

[tex] (1/2)mv_e^2 + (-GMm/r) = 0 + 0 [/tex]

So, we're then left with:

[tex] v_e = \sqrt{2GM/r} [/tex]

Another additional note: I'm not sure why they defined the potential energy the way they did, but that's how they showed it.
 
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  • #6
the "parabola thing" is easy

a small body near an isolated mass must follow an ellipse (eccentricity < 1) a parabola (eccentricity = 1) or a hyperbola (eccentricity > 1)

as you increase the speed (for a fixed direction at a fixed position), you increase the eccentricity

if it's an ellipse, the body will never "reach infinity"

the smallest value of eccentricity, and therefore the smallest speed, for which the body can "reach infinity" is when the path is a parabola :smile:
 

FAQ: Does the angle at which an object is fired affect its escape velocity?

What is escape velocity at an angle?

Escape velocity at an angle refers to the minimum speed at which an object must be launched from a planet's surface at an angle in order to escape its gravitational pull and enter orbit.

How is escape velocity at an angle different from escape velocity?

Escape velocity at an angle takes into account the direction of the launch, whereas escape velocity only considers the vertical launch of an object. This means that an object can achieve escape velocity at an angle with a lower speed than if it were launched straight up.

What factors affect escape velocity at an angle?

The factors that affect escape velocity at an angle include the mass and radius of the planet, as well as the angle and speed of the launch. The larger the planet and the steeper the launch angle, the higher the required escape velocity.

How is escape velocity at an angle calculated?

The formula for calculating escape velocity at an angle is: v = √(2GM/r)(1 + cosθ), where v is the escape velocity, G is the universal gravitational constant, M is the planet's mass, r is the distance from the planet's center, and θ is the launch angle.

Why is escape velocity at an angle important in space travel?

Escape velocity at an angle is important in space travel because it determines the minimum speed at which a spacecraft must launch in order to escape a planet's gravity and enter orbit. It also affects the trajectory and efficiency of the spacecraft's journey. A higher escape velocity at an angle may require more fuel and energy for the spacecraft to reach its destination.

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