INTERESTING KINEMATICS........

2) http://en.wikipedia.org/wiki/Schwarzschild_radius

a) You will deal with this concept of finding minimum velocity with which a object should be projected vertically upward so that it does not returns back to the surface, which is indeed known as Escape velocity in Gravitation.

It is different for different planets. Escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero. It is commonly described as the speed needed to "break free" from a gravitational field

Escape velocity is defined to be the minimum velocity an object must have in order to escape the gravitational field of the earth, that is, escape the earth without ever falling back.

The object must have greater energy than its gravitational binding energy to escape the earth's gravitational field. So:

1/2 mv2 = GMm/R

Where m is the mass of the object, M mass of the earth, G is the gravitational constant, R is the radius of the earth, and v is the escape velocity. It simplifies to:

v = √(2GM/R)

or

v = √(2gR)

Where g is acceleration of gravity on the earth's surface.

The value evaluates to be approximately:

11100 m/s
40200 km/h
25000 miles/h

So, an object which has this velocity at the surface of the earth, will totally escape the earth's gravitational field (ignoring the losses due to the atmosphere.) It is all there is to it.

b) The important point here is that not even light can escape. Therefore the escape velocity of the mass must be greater than the speed of light. The escape velocity is calculated as follow,

1/2 mv² = GMm/R

v² = (2GM/R)

therefore, R = (2GM/v²)

In this case we know v = c = 3 x 108 ms-1, M = 5.97 x 1024 and G = 6.67 x 10^-11m³/kg/s². Put the given values in the above equation & you'll get the answer.

This gives a radius of 8.85 x 10-3 m. This is the maximum radius of a mass equal to that of the earth that is a black hole. A larger radius would give an escape velocity lower than the speed of light, so light to could escape.

By the way, for (c), its not the escape velocity which we require, the orbital velocity (=√gR) will do.

The orbital velocity of the satellite depends on its altitude above Earth. The nearer Earth, the faster the required orbital velocity. At an altitude of 124 miles (200 kilometers), the required orbital velocity is just over 17,000 mph (about 27,400 kph). To maintain an orbit that is 22,223 miles (35,786 km) above Earth, the satellite must orbit at a speed of about 7,000 mph (11,300 kph). That orbital speed and distance permits the satellite to make one revolution in 24 hours. Since Earth also rotates once in 24 hours, a satellite at 22,223 miles altitude stays in a fixed position relative to a point on Earth's surface. Because the satellite stays right over the same spot all the time, this kind of orbit is called "geostationary." Geostationary orbits are ideal for weather satellites and communications satellites.

The moon has an altitude of about 240,000 miles (384,400 km), a velocity of about 2,300 mph (3,700 kph) and its orbit takes 27.322 days. (Note that the moon's orbital velocity is slower because it is farther from Earth than artificial satellites.)

And what I have given is the general case when the satellite is at height h above the surface

of the earth. When the satellite is moving close to the earth surface, i.e, when h<<R, then

we can take (R + h) ≈ R

Hence the above equation becomes

Vo =√GM/R

Also, Vo = √gR {because g = GM/R²}

Vo = v_e/√2,

where v_e = escape velocity of the earth which is 11.2 km/s

By substituting the requisite values in the above equations you can get the orbital velocity

of the satellite, which is 7.9 km/s here.