If the Earth had no atmosphere and were a perfectly smooth sphere, one could launch a projectile into a circular orbit just above the ground. Given that the mass of Earth is 5.98 × 1024 Kg5.98 × 1024 Kg  and its radius is  6.37 × 106 m6.37 × 106 m , how much kinetic energy would be required to send a 2 km ball into such an orbit?

To solve this problem, we need to use the formula for the kinetic energy of an object in orbit, which is given by:

KE = 1/2 * m * v^2

where m is the mass of the object and v is its velocity. The velocity of an object in orbit is given by the formula:

v = sqrt(G * M / r)

where G is the gravitational constant (6.674 * 10^-11 m^3 kg^-1 s^-2), M is the mass of the Earth, and r is the radius of the Earth plus the radius of the object (since the object is in orbit just above the ground).

First, we need to calculate the velocity of the object. Given that the mass of the Earth is 5.98 * 10^24 kg and its radius is 6.37 * 10^6 m, and the radius of the ball is 1 km (since the diameter is 2 km), we have:

v = sqrt((6.674 * 10^-11 m^3 kg^-1 s^-2) * (5.98 * 10^24 kg) / ((6.37 * 10^6 m) + (1 * 10^3 m))) v = sqrt((6.674 * 10^-11 m^3 kg^-1 s^-2) * (5.98 * 10^24 kg) / (6.371 * 10^6 m)) v = 7.91 * 10^3 m/s

Then, we can calculate the kinetic energy of the object. Given that the mass of the object is 2 km * pi * r^2 * density (assuming the ball is made of rock with a density of about 3000 kg/m^3), we have:

m = 2 * 10^3 m * pi * (1 * 10^3 m)^2 * 3000 kg/m^3 m = 6.28 * 10^9 kg

KE = 1/2 * (6.28 * 10^9 kg) * (7.91 * 10^3 m/s)^2 KE = 1.98 * 10^17 J

So, the kinetic energy required to send a 2 km ball into such an orbit is approximately 1.98 * 10^17 Joules.