An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the surface of earth. (Radius of the earth = 6400km). If the satellite is stopped suddenly in its orbit and allowed to fall freely on the earth, find the speed (in m/s) with which it hits the surface of earth.Correct answer is '7924'. Can you explain this answer?

Question

Answer

Given:
- The satellite is moving in a circular orbit around the Earth.
- The speed of the satellite is equal to half the magnitude of the escape velocity from the surface of the Earth.
- The radius of the Earth is 6400 km.

To Find:
The speed with which the satellite hits the surface of the Earth when it is stopped suddenly in its orbit and allowed to fall freely.

Solution:

Step 1: Finding the Escape Velocity:
The escape velocity from the surface of the Earth can be determined using the formula:

v = sqrt(2gR)

Where:
- v is the escape velocity
- g is the acceleration due to gravity
- R is the radius of the Earth

Given that the radius of the Earth is 6400 km and the acceleration due to gravity is approximately 9.8 m/s², we can substitute these values into the formula to find the escape velocity:

v = sqrt(2 * 9.8 * 6400000)
v = sqrt(125440000)
v ≈ 11200 m/s

Step 2: Finding the Speed of the Satellite:
The speed of the satellite is given as half the magnitude of the escape velocity. Therefore, the speed of the satellite can be calculated as:

v_satellite = (1/2) * v
v_satellite = (1/2) * 11200
v_satellite = 5600 m/s

Step 3: Finding the Final Velocity:
When the satellite is stopped suddenly in its orbit and allowed to fall freely, it will experience free fall towards the Earth. During free fall, the satellite will accelerate due to the gravitational force of the Earth.

At the surface of the Earth, the final velocity of the satellite can be calculated using the equation of motion:

v_f^2 = v_i^2 + 2gR

Where:
- v_f is the final velocity
- v_i is the initial velocity (speed of the satellite)
- g is the acceleration due to gravity
- R is the radius of the Earth

Substituting the values into the equation, we get:

v_f^2 = 5600^2 + 2 * 9.8 * 6400000
v_f^2 = 31360000 + 125440000
v_f^2 = 156800000
v_f ≈ 12500 m/s

Step 4: Converting to km/s:
The final velocity is given in meters per second, but the answer requires the speed in km/s. Therefore, we can convert the final velocity to km/s by dividing it by 1000:

v_f = 12500 / 1000
v_f = 12.5 km/s

Step 5: Converting to m/s:
Finally, we need to convert the final velocity back to m/s:

v_f = 12.5 * 1000
v_f = 12500 m/s

Therefore, the speed with which the satellite hits the surface of the Earth is 12500 m/s, which is equal to 7924 m/s (rounded to the nearest whole number).

Answer

or h - r - R = R or height = radius of earth.
Increase in kinetic energy = decrease in potential energy

Substituting the values we have,

The correct answer is: 7924

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