Earth escape velocity vs escape speed

By tchall, September 7, 2009 10:12 AM

Escape Velocity and Escape Speed

Escape Velocity and Escape Speed are two similar yet different terms.
This surface escape velocity is the speed required for an object to leave a planet if the object is simply projected from the surface of a planet and then left without any more kinetic energy input.
In practice the vehicle’s propulsion system will continue to provide energy after it has left the surface.

A planet has mass M (Earth 6.0×1024 kilograms), and a planet has gravity and the object is located a distance from the center of the planet or planets radius r (Earth’s radius is 6.4×106 meters) and the object you are trying to project has a mass m.

Thus

ve = EQUATION GOES HERE

Escape Velocity

Newtons example of escape velocity of a projectile

Newtons example of escape velocity of a projectile

This more or less refers to projectiles, and the initial speed required of them to escape the gravitational forces of a planet.
Take for example a rile and if you were to fire it upwards, what would be the required initial velocity required of the bullet to escape the earths gravity.
i.e. so that gravity will never manage to pull it back.

This would depend on the mass of the planet and the distance from the center point.

On the surface of the Earth, the escape velocity is about 11.2 kilometers per second, 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.

If an object moves fast enough it can escape a massive object’s gravity and not be drawn back toward the massive object.

More specifically, this is the initial speed something needs to escape the object’s gravity and assumes that there is no other force acting on the object besides gravity after the initial boost.

This is not the case with rockets.  Their intial speed is 0 km/s, and then this is gradually accelerated as they continue to thrust upwards.

Rockets leaving the Earth do not have the escape velocity at the beginning but the engines provide thrust for an extended period of time, so the rockets can eventually escape. The concept of escape velocity applies to anything gravitationally attracted to anything else (gas particles in planet atmospheres, comets orbiting the Sun, light trying to escape from black holes, galaxies orbiting each other, etc.).

How do you do that?

Find the escape velocity from the surface of the Earth. Using the acceleration of gravity, you can find that the Earth has a mass of 6.0×1024 kilograms. The Earth’s radius is 6.4×10 6 meters. Since the mass and distance from the center are in the standard units, you just need to plug their values into the escape velocity relation.
The Earth’s surface escape velocity is Sqrt[2× (6.7×10-11) × (6.0×10 24)/ (6.4×10 6)] = Sqrt[1.256×10 8] = 1.1×104 meters/second (= 11 km/s).

Here are some other surface escape velocities: Moon = 2.4 km/s, Jupiter = 59.6 km/s, Sun = 618 km/s.

Formula
Mass of central object = [(orbital speed)2 × distance)/G.
Mass of central object (Kepler's 3rd law) = (4p2)/G × [(distance)3/(orbital period)2].
Orbital speed = Sqrt[G × Mass / distance].
Escape velocity = Sqrt[2G × Mass / distance].

Escape velocity relative to Equator and Spin of the Earth

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).

Gravitational Space Wells

Gravitational Space Wells

Leave a Reply

After Hours Doctors After Hours Doctors