What is the condition for escape? (hint: what is the potential energy and what is the kinetic energy of the body when it has reached a distance where the Earth's gravity is negligible and the body is just barely moving?). Use the principle of conservation of total energy to determine what its speed must be initially in order to achieve that distance.Math10 said:The Attempt at a Solution
m(dv/dt)=-(mgR^2)/(x+R)^2
dv/dt=-(gR^2)/(x+R)^2
Now what?
Escape velocity is the minimum speed an object needs to reach in order to break free from the gravitational pull of a celestial body, such as a planet or moon.
The escape velocity can be calculated using the formula: v = √(2GM/r), where v is the escape velocity, 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's starting point.
The unit of escape velocity is typically meters per second (m/s). However, it can also be expressed in other units such as kilometers per second (km/s) or miles per second (mi/s).
The escape velocity depends on the mass and radius of the celestial body. Generally, larger and more massive bodies have higher escape velocities compared to smaller and less massive bodies. For example, the escape velocity on Earth is 11.2 km/s, while on the Moon it is only 2.4 km/s.
The escape velocity of Earth is approximately 6.9 miles per second (mi/s). However, this value may vary slightly depending on the altitude and location on Earth.