Calculate the velocity of a 20kg object that is dropped from a height h above the Earth, where h is very large compared to the radius of the Earth. The radius of the Earth is 6380 m and the mass is 5.94 x 10^24 kg.
I would do this as energy.
1/2 m v^2=work done=ind force*dh=INT GMe*m/(re+h)^2 dh integrating from intial height to Re. Then solve for v.
To calculate the velocity of an object that is dropped from a height h above the Earth, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system is the sum of potential energy and kinetic energy.
At the start, when the object is at height h, it has potential energy (PE) which can be calculated using the formula:
PE = mgh
where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height above the surface of the Earth.
When the object falls and reaches the Earth's surface, all the potential energy is converted into kinetic energy (KE). The kinetic energy is given by the formula:
KE = 1/2mv²
where v is the velocity of the object.
Using the principle of conservation of mechanical energy, we can equate the initial potential energy to the final kinetic energy:
mgh = 1/2mv²
Dividing both sides of the equation by m and canceling the 'm' term, we get:
gh = 1/2v²
Simplifying further, we have:
v² = 2gh
To get the velocity, we can simply take the square root of both sides of the equation:
v = √(2gh)
Now, substituting the given values:
m = 20 kg
h = very large compared to the radius of the Earth (approximately ∞)
r = radius of the Earth = 6380 m
M = mass of the Earth = 5.94 x 10^24 kg
g = gravitational acceleration = GM/r² = (6.67 x 10^-11 m³/(kg s²)) * (5.94 x 10^24 kg) / (6380 m)²
Calculating the value of g and substituting it back into the velocity equation, we can find the velocity of the object.