If Superman could throw a baseball very fast and air friction could be ignored, he could put the baseball into orbit. Assume that the balls orbital radius equals 6.38x10^6m. Earth has a mass of 5.98x10^24kg. How fast must Superman throw the ball?
recall that escape velocity
v = √(2GM/d)
2gM
To determine the velocity at which Superman must throw the baseball in order for it to enter orbit, we can make use of the principle of conservation of energy.
The total mechanical energy of the baseball in orbit is given by the sum of its kinetic energy (KE) and the gravitational potential energy (PE) at that point:
Total mechanical energy = KE + PE
The kinetic energy of the baseball is given by the equation KE = (1/2)mv^2, where m is the mass of the baseball and v is its velocity.
The potential energy of the baseball is given by the equation PE = -G * (m * M) / r, where G is the gravitational constant, M is the mass of the Earth, and r is the radius of the orbit.
At the point of orbit, the total mechanical energy is equal to zero (since the ball is in a stable orbit). Therefore, we can write:
0 = (1/2)mv^2 - G * (m * M) / r
Rearranging the equation to solve for v, the velocity of the baseball:
(1/2)mv^2 = G * (m * M) / r
v^2 = 2 * G * M / r
v = sqrt(2 * G * M / r)
Now, we can substitute the given values:
Gravitational constant, G = 6.67430 x 10^(-11) N m^2 / kg^2
Mass of the Earth, M = 5.98 x 10^24 kg
Radius of the orbit, r = 6.38 x 10^6 m
Plugging in these values, we can calculate the velocity:
v = sqrt(2 * 6.67430 x 10^(-11) N m^2 / kg^2 * 5.98 x 10^24 kg / 6.38 x 10^6 m)
Calculating this, we find:
v ≈ 7.85 x 10^3 m/s
Therefore, Superman must throw the ball with a velocity of approximately 7.85 x 10^3 m/s in order for it to enter orbit.
To determine the speed at which Superman must throw the baseball in order to put it into orbit, we can use the principle of conservation of energy. The total mechanical energy of the baseball in orbit will be equal to the sum of its kinetic energy and gravitational potential energy.
The kinetic energy (KE) of an object is given by the equation:
KE = (1/2)mv^2
where m is the mass of the object and v is its velocity.
The gravitational potential energy (PE) of an object in orbit is given by the equation:
PE = (-GMm) / r
where G is the gravitational constant, M is the mass of Earth, m is the mass of the object, and r is the orbital radius.
In this case, the baseball is being thrown by Superman, so the mass of the baseball can be denoted as m, and the mass of Earth as M.
Setting the total mechanical energy equal to zero gives us:
KE + PE = 0
(1/2)mv^2 + (-GMm / r) = 0
Since r is given in the problem as the orbital radius, we can substitute the value of r:
(1/2)mv^2 + (-GMm / 6.38x10^6) = 0
Using the given values, M = 5.98x10^24 kg and r = 6.38x10^6 m, and assuming that the mass of the baseball (m) is negligible compared to the mass of Earth, we can simplify the equation:
(1/2)v^2 = (GM / r)
To solve for the velocity (v), we can rearrange the equation:
v^2 = (2GM / r)
v = √(2GM / r)
Substituting the values of G, M, and r:
v = √(2(6.67x10^-11 Nm^2/kg^2)(5.98x10^24 kg) / (6.38x10^6 m))
Simplifying and calculating the equation will give us the required speed at which Superman must throw the baseball to put it into orbit around Earth.