OMGosh - somebody shoot me...

Paula K.
on 7/8/09 10:58 am - Laguna Hills, CA

Or at least tell me how to figure out the Escape Velocity of Earth......

My head is going to explode before I figure out this thing.

Blech.

39 lbs lost prior to surgery!

rlfroo
on 7/8/09 11:26 am - Hesperia, CA

RNY on 08/01/08

I am sorry Paula, but I don't have a clue what you are talking about. I would love to help, but I just don't know. You need to understand I am old and the brain does not work so if you are speaking in code, I just don't get it.

Rhonda

Paula K.
on 7/8/09 11:48 am - Laguna Hills, CA

RNY on 02/23/09

No code here, Rhonda. Just pure, unadulterated confusion =0)

39 lbs lost prior to surgery!

rlfroo
on 7/8/09 11:51 am - Hesperia, CA

RNY on 08/01/08

I am sorry you are confused. How can we help? PM me if you wish we have a group here and we could bring it up as our evening topic if you would like and we will try our best to un confuse you.

Rhonda

Darlene
on 7/8/09 11:39 am

RNY on 08/26/00

To leave planet Earth an escape velocity of 11.2 km/s is required, however a speed of 42.1 km/s is required to escape the Sun's gravity (and exit the solar system) from the same position

Women are angels.
...and when someone breaks our wings, we simply continue to fly...on a broomstick.

We are flexible.

Darlene

rlfroo
on 7/8/09 11:43 am - Hesperia, CA

RNY on 08/01/08

Darelene,

Thanks for clearing that up for us.

Appriciativley,

Rhonda

cabin111
on 7/8/09 12:38 pm

RNY on 10/09/06

I'm terrible at math. I got this from the search engine "bing" and "wiki". Wonder if these will help below or make it more confusing. Good luck, Brian
For other senses of this term, see escape velocity (disambiguation).

Isaac Newton's analysis of escape velocity. Projectiles A and B fall back to earth. Projectile C achieves a circular orbit, D an elliptical one. Projectile E escapes.

In physics, escape velocity is the speed where the kinetic energy of an object is equal to the magnitude of its gravitational potential energy, as calculated by the equation,

$U_g = \frac{-Gm_1m_2}{r}$

It is commonly described as the speed needed to "break free" from a gravitational field (without any additional impulse) and is theoretical, totally neglecting atmospheric friction. The term escape velocity can be considered a misnomer because it is actually a speed rather than a velocity, i.e. it specifies how fast the object must move but the direction of movement is irrelevant, unless "downward." In more technical terms, escape velocity is a scalar (and not a vector).

Escape velocity gives a minimum delta-v budget for rockets when no benefit can be obtained from the speeds of other bodies for a particular mission; but it neglects losses such as air drag and gravity drag. However in some cases it can be improved upon, for example, by use of gravitational slingshots.

[edit] Overview

The phenomenon of escape velocity is a consequence of conservation of energy. For an object with a given total energy, which is moving subject to conservative forces (such as a static gravity field) it is only possible for the object to reach combinations of places and speeds which have that total energy; and places which have a higher potential energy than this cannot be reached at all.

For a given gravitational potential energy at a given position, the escape velocity is the minimum speed an object without propulsion needs to have sufficient energy to be able to "escape" from the gravity, i.e. so that gravity will never manage to pull it back. For the sake of simplicity, unless stated otherwise, we will assume that the scenario we are dealing with is that an object is attempting to escape from a uniform spherical planet by moving straight up (along a radial line away from the center of the planet), and that the only significant force acting on the moving object is the planet's gravity.

Escape velocity is actually a speed (not a velocity) because it does not specify a direction: no matter what the direction of travel is, the object can escape the gravitational field (though its path may intersect the earth). The simplest way of deriving the formula for escape velocity is to use conservation of energy. Imagine that a spaceship of mass m is at a distance r from the center of mass of the planet, whose mass is M. Its initial speed is equal to its escape velocity, $v e$ . At its final state, it will be an infinite distance away from the planet, and its speed will be negligibly small and assumed to be 0. Kinetic energy K and gravitational potential energy U_g are the only types of energy that we will deal with, so by the conservation of 'energy',

$(K + U_g)_i = (K + U_g)_f \,$

K_ƒ = 0 because final velocity is zero, and U_gƒ = 0 because its final distance is infinity, so

$\frac{1}{2}mv_e^2 + \frac{-GMm}{r} = 0 + 0$

$v_e = \sqrt{\frac{2GM}{r}}$

Defined a little more formally, "escape velocity" is the initial speed required to go from an initial point in a gravitational potential field to infinity with a residual velocity of zero, with all speeds and velocities measured with respect to the field. Additionally, the escape velocity at a point in space is equal to the speed that an object would have if it started at rest from an infinite distance and was pulled by gravity to that point. In common usage, the initial point is on the surface of a planet or moon. On the surface of the Earth, the escape velocity is about 11.2 kilometers per second (~6.96 mi/s), which is approximately 34 times the speed of sound (mach 34) and at least 10 times the speed of a rifle bullet. However, at 9,000 km altitude in "space", it is slightly less than 7.1 km/s.

The escape velocity relative to the surface of a rotating body depends on direction in which the escaping body travels. For example, as the Earth's rotational velocity is 465 m/s at the equator, a rocket launched tangentially from the Earth's equator to the east requires an initial velocity of about 10.735 km/s relative to Earth to escape whereas a rocket launched tangentially from the Earth's equator to the west requires an initial velocity of about 11.665 km/s relative to Earth. The surface velocity decreases with the cosine of the geographic latitude, so space launch facilities are often located as close to the equator as feasible, e.g. the American Cape Canaveral (latitude 28°28' N) and the French Guiana Space Centre (latitude 5°14' N).

Paula K.
on 7/8/09 2:52 pm - Laguna Hills, CA

RNY on 02/23/09

My head hurts.

But thank you!

39 lbs lost prior to surgery!

Jackie717
on 7/8/09 2:50 pm

RNY on 07/16/09

OMG now Im confused!

Started Liquid pre-op diet July 2, 2009 / Surgery on July 16th, 2009
Starting weight: 385
Current: 245
First goal: 180

cabin111
on 7/8/09 3:17 pm

RNY on 10/09/06

Maybe this will make it easier...
That amount is the integral of the expression above:

$\begin{align} \int_r^\infty gm \left(\frac{r}{s}\right)^2 \, ds & = gmr^2 \int_r^\infty s^{-2}\,ds = gmr^2 \left[-s^{-1}\right]_{s:=r}^{s:=\infty} \\ & = gmr^2\left(0-(-r^{-1})\right)=gmr. \end{align}$

That is how much kinetic energy the object of mass m needs in order to escape. The kinetic energy of an object of mass m moving at speed v is (1/2)mv². Thus we need

$\begin{matrix}\frac12\end{matrix} mv^2=gmr.$