Answer
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Hint: The escape velocity is a velocity that is required by an object in order to overcome the gravitational pull of a body (like earth). The kinetic energy of the object has to be more than (or equal to) the gravitational potential of the earth.
Formula used:
Escape velocity is given by
$v= \sqrt{\dfrac{2GM}{r}}$
r is the distance from the centre of the earth.
Complete answer:
Consider an object at a height of h from the earth surface. This means that it is at a distance of r=R+h from the center of the earth.
The escape velocity is determined by equating the kinetic energy of the particle with the gravitational potential it experiences due to earth. If the kinetic energy is more than the gravitational pull, then certainly the object will escape the earth's field.
We know that the gravitational potential has a magnitude given by:
$\dfrac{GMm}{r}$
And kinetic energy as usual is
$\dfrac{1}{2} mv^2 $
So, equating the two gives us:
$\dfrac{1}{2} mv^2 = \dfrac{GMm}{r}$
This gives us the escape velocity:
$v= \sqrt{\dfrac{2GM}{r}}$
We are given the value of g and we know g is expressed as:
$g= \dfrac{GM}{R^2}$
Therefore we modify the escape velocity formula as:
$v= \sqrt{\dfrac{2GM R^2}{r R^2}}$
which upon keeping g, can be simply written as:
$v= \sqrt{\dfrac{2g R^2}{r}}$
Keeping r=R+h and substituting the given values i.e., h = 1000 km, R = 6400 km and g = 9.8 $m/s^2$;
$v= \sqrt{\dfrac{2 \times 9.8 \times (6400000)^2}{7400000}}$
This gives the velocity to be about
v= 10415.79 m/s
Therefore the correct answer is 10.4 km/s.
So, the correct answer is “Option C”.
Note:
The formula for escape velocity should contain a r in place of R because our object is at a distance. One might not regard this and simply use R. In such situations, one can confirm the formula by equating Kinetic and potential energy. Gravitational potential becomes weaker and weaker as we move farther and farther away from the earth. Another common mistake is putting distance in km in the square root. One must convert the km into m first.
Formula used:
Escape velocity is given by
$v= \sqrt{\dfrac{2GM}{r}}$
r is the distance from the centre of the earth.
Complete answer:
Consider an object at a height of h from the earth surface. This means that it is at a distance of r=R+h from the center of the earth.
The escape velocity is determined by equating the kinetic energy of the particle with the gravitational potential it experiences due to earth. If the kinetic energy is more than the gravitational pull, then certainly the object will escape the earth's field.
We know that the gravitational potential has a magnitude given by:
$\dfrac{GMm}{r}$
And kinetic energy as usual is
$\dfrac{1}{2} mv^2 $
So, equating the two gives us:
$\dfrac{1}{2} mv^2 = \dfrac{GMm}{r}$
This gives us the escape velocity:
$v= \sqrt{\dfrac{2GM}{r}}$
We are given the value of g and we know g is expressed as:
$g= \dfrac{GM}{R^2}$
Therefore we modify the escape velocity formula as:
$v= \sqrt{\dfrac{2GM R^2}{r R^2}}$
which upon keeping g, can be simply written as:
$v= \sqrt{\dfrac{2g R^2}{r}}$
Keeping r=R+h and substituting the given values i.e., h = 1000 km, R = 6400 km and g = 9.8 $m/s^2$;
$v= \sqrt{\dfrac{2 \times 9.8 \times (6400000)^2}{7400000}}$
This gives the velocity to be about
v= 10415.79 m/s
Therefore the correct answer is 10.4 km/s.
So, the correct answer is “Option C”.
Note:
The formula for escape velocity should contain a r in place of R because our object is at a distance. One might not regard this and simply use R. In such situations, one can confirm the formula by equating Kinetic and potential energy. Gravitational potential becomes weaker and weaker as we move farther and farther away from the earth. Another common mistake is putting distance in km in the square root. One must convert the km into m first.
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