A rocket is launched straight up from the earth's surface at a speed of 1.80×104 m/s .

Asked by maham237 @ 15/12/2020 in Physics viewed by 455 People

part awhat is its speed when it is very far away from the earth?

Answered by maham237 @ 15/12/2020

We can solve the problem by using the law of conservation of energy.

When the rocket starts its motion from the Earth surface, its mechanical energy is sum of kinetic energy and gravitational potential energy:
$E_i = K_i + U_i = \frac{1}{2} m v_i^2 + (- \frac{GM}{r} )$
where
m is the rocket's mass
$v_i = 1.8 \cdot 10^4 m/s$ is the rocket initial speed
$G=6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant
$M=5.97 \cdot 10^{24} kg$ is the Earth's mass
$r= 6.37 \cdot 10^6 m$ is the distance of the rocket from the Earth's center (so, it corresponds to the Earth's radius)

The mechanical energy of the rocket when it is very far from the Earth is just kinetic energy (because the gravitational potential at infinite distance from Earth is taken to be zero):
$E_f = K_f = \frac{1}{2} mv_f ^2$
where $v_f$ is the final speed of the rocket.

By equalizing the initial energy and the final energy, we can find the final velocity:
$\frac{1}{2} mv_i ^2 - \frac{GM}{r} = \frac{1}{2}m v_f^2$
$v_f = \sqrt{v_i^2 - \frac{GM}{r} } =1.41 \cdot 10^4 m/s$

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A rocket is launched straight up from the earth's surface at a speed of 1.80×104 m/s .

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