Then set the initial KEnergy equal to the potential energy at the surface of EArth.
1/2 mv^2= GMe*m/re
solve for escape velocity, v
The gravitational constant for g is, G = 6.67428*10^-11 m^3 Kg^-1 s^-2
Please answer the question fully and please give any helpful websites, thankyou.
Also I know calculus.
1/2 mv^2= GMe*m/re
solve for escape velocity, v
Is that right
1/2 m v^2=GMe m/re
v^2= 2GMe/re
take the square root of each side.
The escape velocity can be calculated using the following formula:
v = sqrt((2 * G * M) / r)
Where:
- v is the escape velocity
- G is the gravitational constant (6.67428 x 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the Earth (approximately 5.972 x 10^24 kg)
- r is the distance from the center of the Earth to the object
In this case, since we want to find the initial velocity for a projectile to leave the Earth's surface, the distance r will be the radius of the Earth (approximately 6.371 x 10^6 m).
So, plugging in the values:
v = sqrt((2 * G * M) / r)
v = sqrt((2 * 6.67428 x 10^-11 m^3 kg^-1 s^-2 * 5.972 x 10^24 kg) / (6.371 x 10^6 m))
By evaluating this expression, you will get the magnitude of the initial velocity that a projectile must possess to leave the Earth when air friction is neglected.
If you would like to explore more about the law of gravitation or escape velocity, you can refer to the following helpful websites:
1. NASA's website on the law of gravitation: https://www.grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/gravity_force_calc.htm
2. Khan Academy's topic on escape velocity: https://www.khanacademy.org/science/physics/centripetal-force-and-gravitation/gravity-newtonian/v/escape-velocity