Motion Physics Examples Solved



Consider a simple model for a rocket launched from the surface of the Earth. A better expression for a rocket’s position measured from the center of the Earth is given by:


where RE is the radius of the Earth (6.38 ✕ 106m) and g is the constant acceleration
of an object in free fall near the Earth’s surface (9.81 m/s2).

(a) Determine expressions for velocity and acceleration of the rocket
(b) When will the rocket be at y = 4R_{E} ?
(c) What are the velocity and the acceleration when y = 4R_{E} ?


(a) Velocity of the rocket is


Acceleration of the rocket is


(b) When y = 4R_{E}


(c) The velocity of the rocket when y = 4R_{E} or \frac{7}{3}\sqrt{\frac{2R_{E}}{g}} is

v_y = \frac{R_E\sqrt{2g}}{(R_E^{3/2} + 3\sqrt{\frac{g}{2}}R_E * \frac{7}{3} \sqrt{\frac{2R_{E}}{g}})^{1/3}} = \frac{R_E\sqrt{2g}}{(8R_{E}^{3/2})^{1/3}} = \sqrt{\frac{gR_{E}}{2}} (m/s)

The acceleration of the rocket when y = 4R_{E} or t = \frac{7}{3} \sqrt{\frac{2R_{E}}{g}} is



(a) 06

(b) 07

(c) 08

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