It is inherent in the equations of motion. The object does slow down some but not down to zero. That's really the definition of escape velocity; giving an object sufficient velocity that gravity never quite overcomes it. Mathematically, when it gets to infinity it has zero velocity relative to the Earth (assuming it left with exactly the escape velocity), but of course in reality it will be more affected by celestial bodies other than the Earth WAY long before then (and the infinite "then" never actually occurs in reality)Zynoakib said:I know once the escape velocity is reached, the object will continuous to move away from the Earth. But the Earth's gravity can still act on the object no matter how far it goes, so what keeps the object from stopping or even returning back to Earth?
Thanks in advance!
phinds said:It is inherent in the equations of motion. The object does slow down some but not down to zero. That's really the definition of escape velocity; giving an object sufficient velocity that gravity never quite overcomes it. Mathematically, when it gets to infinity it has zero velocity relative to the Earth (assuming it left with exactly the escape velocity), but of course in reality it will be more affected by celestial bodies other than the Earth WAY long before then (and the infinite "then" never actually occurs in reality)
No, that's not a good way to say it. It will take forever for it to be slowed to zero.Zynoakib said:Just want to explain it in my own wording
Although the Earth's gravity can still attract the object to slow it down no matter how far it goes, it will take forever for the object to be slowed down just by a slight degree.
So, note that escape velocity is not a single/constant value. It decreases with distance. So an object that is launched exactly at escape velocity will always be slowing down, and will always be at the escape velocity of whatever distance it is at.phinds said:No, that's not a good way to say it. It will take forever for it to be slowed to zero.
Agreed for sure.russ_watters said:So, note that escape velocity is not a single/constant value. It decreases with distance. So an object that is launched exactly at escape velocity will always be slowing down, and will always be at the escape velocity of whatever distance it is at.
The gravitational pull of Earth is the force exerted by the Earth's mass on objects near its surface. This force is what keeps objects, including humans, on the surface of the Earth and causes objects to fall towards the ground.
The gravitational pull of Earth can be calculated using Newton's law of universal gravitation, which states that the force of gravity between two objects is directly proportional to their masses and inversely proportional to the square of the distance between them. The equation for calculating the gravitational pull of Earth is F = G * (m1 * m2)/r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.
Escape velocity is the minimum speed that an object needs to achieve in order to escape the gravitational pull of a celestial body, such as Earth. It is the speed at which the object's kinetic energy is greater than the gravitational potential energy of the body it is trying to escape from.
The escape velocity of a celestial body, such as Earth, can be calculated using the equation v = √(2GM/r), where G is the gravitational constant, M is the mass of the celestial body, and r is the distance from the center of the body to the object. This equation assumes that the object is starting at rest at an infinite distance from the body.
The escape velocity of Earth is approximately 11.2 kilometers per second (km/s) or 6.95 miles per second (mi/s). This means that in order for an object to escape the gravitational pull of Earth, it would need to reach a speed of 11.2 km/s or 6.95 mi/s.