Question

Newton's constant of gravitation G is 6.67×10^-11 in the system of units we use with mass in kilograms, length in meters, and time in seconds. The radius of Earth is approximately 6.378 ×10⁶ m, and its mass is 5.97 × 10²⁴ kg. With all this you can evaluate conservation of energy from Earth's surface and find an escape velocity of about 11.2 km/s from

Now suppose we wanted to orbit a spacecraft in low Earth orbit just a few hundred km above the surface. How fast would it have to go to stay in a circular orbit?

a)		7.9 km/s, that is about 11.2 / √2.
b)		5.6, half the escape velocity
c)		11.2, the same as the escape velocity
d)		15.8, that is 11.2 * √2.

Answer 1

This is how escape velocity defined.
at r-> oo Escape speed at earth's surface: Suppose a particle of mass m is on earth's surface We project it with a velocity V from the earth's surface, so that it just reaches r → (at r → 00, its velocity become zero) Applying energy conservation between initial position (when the particle was at earth's surface) and find positions (when the particle just reaches Earth M., R GMe 2 2GM 2GM Escape speed from earth is surface Ve If we put the values of G, Me, R the we get Ve= 11.2 km/s

In simple term if we project any object with this velocity it will never return.

If we want to fix a satellite. we need orbital velocity by which any satellite moves around the earth this we need

CIRCULAR MOTION OF A SATELLITE AROUND A PLANET Planetナーr·--mo --… Suppose at satellite of mass mo is at a distance r from a planet. If the satellite does not revolve, then due to the gravitational attraction, it may collide to the planet.

r=R_earth +few hundreds KM neglect  few hundreds KM with R_earth  

To avoid the collision, the satellite revolve around the planet, for circular motion of satellite moV..1) GMe this velocity is called orbital velocity (Vo) GM 0

We get $V_{0}=\sqrt{GM_{e}/R_{e}}$

$V_{e}=\sqrt{2}V_{0}$

Thats why A is correct