plz solve the question for me,,,,,,,,,,,,,,,,,,PLZ?

Question

an artificial satellite is moving in a circular orbit with spd equal to half the escape velocity from earth's surface.If the satellite is stopped suddenly and is allowed to fall freely onto the earth,what is the speed with which it strikes the surface of the earth

Answer 1

To find escape velocity :

m*v^2 / 2 = G*M*m / R

m = mass of the satellite

M = mass of the planet

G = constant

R = radius of the planet

v = escape velocity

m*v^2 / 2 = GMm / R

v = sqrt(2GM / R)

v/2 = 1/2*sqrt(2GM / R)

the kinetic plus the potential energy of the satellite :

m*2GM/R / 8 - GMm / (R + h)

energy when it falls : m*v^2 / 2 - GMm / R

let's consider h = 0

2GMm / 8R - GMm / R = mv^2 / 2 - GMm / R

v = sqrt( GM / 2R)

where :

M = mass of the planet

R = radius

G = gravity constant

this is the answer, if you consider h = 0, and that's not true.

If we not consider it :

2GMm / 8R - GMm / (R+h) = mv^2 / 2 - GMm / R

and if it's circular orbit : m*v^2 / (R+h) = GMm / (R + h)^2

the centripetal force is equal to the gravitational force :

then you have V : 1/2*sqrt(2GM / R) >>> speed of the satellite

then : 2GM / 4R(R+h) = GMm / ( R + h) ^2

then : R = h

we consider :

2GMm / 8R - GMm / (R+h) = mv^2 / 2 - GMm / R

R = h

2GMm / 8R - GMm / 2R = mv^2 / 2 - 2GMm / 2R

2GMm / 8R - 4GMm / 8r = mv^2 / 2 - 8GMm / 8R

mv^2 / 2 = 6GMm / 8R

v = sqrt(3GM / 2) m/s

The answers will be :

for h = 0 : v = sqrt( GM / 2R) m /s

for h = R : v = sqrt(3GM / 2R) m/s