Projectiles and escape velocity [closed]

Q: The escape velocity for a body projected vertically upwards from the surface of earth is 11 km/s. If the body is projected at an angle of $45^\circ$ with vertical, the escape velocity will be?

My Approach:

The new vertical velocity will be $u * \sin(45^\circ)$ = $u /\sqrt2$

Calculating expression for escape velocity:

$$\frac{1}{2} mv^2 = \frac{GMm}{r}$$ $$\frac{1}{4} mu^2 = \frac{GMm}{r}$$ $$u = 2\sqrt\frac{GMm}{r}$$

Hence the new escape velocity = $\sqrt2 * 11$km/s.

However, the Correct Answer is 11km/s.

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closed as off-topic by Colin McFaul, Kyle Kanos, Brandon Enright, Dan, Chris WhiteJun 11 '14 at 20:13

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Escape velocity is not dependent on direction. It really should be called "escape speed"

It's the result of the calculation "kinetic energy of object at launch"="potential energy change on leaving Earth's influence"

Kinetic energy isn't a vector. It is $\frac12 m\vec v \cdot \vec v=\frac12 m|\vec v|^2$; a scalar quantity. Note that it is independsnt of the direction of the velocity, only the vector.

Remember, if you throw a ball at 45 degrees, the Earth's gravity will attract it, but it will also speed it up. These two effects "cancel out"

Fun fact: If there was a tunnel through the Earth, and you threw a ball at escape velocity down the tunnel, it would still escape Earth's influence after passing through the tunnel.

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Well, this certainly is an evil trick to play on first year students!

Escape velocity isn't actually a velocity at all. It's a speed, i.e., it's scalar quantity as opposed to a vector quantity. Note that when the escape "velocity" at r was calculated, the only assumption made was conservation of mechanical energy, and then magnitude of v is isolated from kinetic energy. No direction is presumed, and so it applies any direction.

Hence, v = 11km/s, projected vertically or otherwise.

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Yes, or more to the point the kinetic energy isn't a vector. So long as the kinetic energy is equal to the gravitational potential energy the object will escape regardless of what direction it's moving in (unless of course it points down and isn't a neutrino :-). – John Rennie Apr 9 '13 at 9:11
Adding to the evilness is that in the real world Aneesh Dogra's intuition that the numbers should be different is correct because launching at an angle will result in more deceleration from air drag before clearing the atmosphere. – Dan Neely Apr 9 '13 at 15:37
@DanNeely While that's worth pointing out, I would counter by saying "Who cares?" =p Because the escape velocity three times a mathematical idealization: a theoretical minimum value from which a body ends up with exactly zero kinetic energy at infinity. Obviously it's intended purpose is more for providing a natural reference for scaling and analyzing the equations of motion. No one ever made direct use of escape velocity when it came down to actual implementation. – David H Apr 9 '13 at 15:52

The (relative) escape velocity of a mass $M$ with respect to a mass $m$ is: $$V_{rel} = \sqrt{\frac{2 G(M\!+\!m)}{r} }$$where $r$ is the relative distance of the two bodies.

The escape velocity of a mass $m$ with respect to the inertial center of mass of the two bodies is: $$V_m = \sqrt{ \frac{2 G M M}{ r\, (M\!+\!m)} }$$ and the escape velocity of the mass $M$ with respect to the inertial center of mass is: $$V_M = \sqrt{ \frac{2 G m m}{ r\, (M\!+\!m)} }$$

The relative velocity is the sum of the two inertial velocities: $V_{rel} = V_M + V_m$.

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