Question

A rocket is launched straight up from the earth's surface at a speed of 1.80×104 m/s .part awhat is its speed when it is very far away from the earth?

Answer 1

The speed of the rocket when it is very far away from the Earth is  $\boxed{1.61 \times {{10}^4}\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}}$.

Further Explanation:

Since the gravitational potential energy of the rocket will become when the rocket reaches a distance far away from the Earth, the complete kinetic and gravitational potential energy of the rocket on the surface of Earth is converted into the kinetic energy of the rocket.

The total energy of the rocket on the surface of the Earth is:

$T{E_i} = {K_i} + {U_i}$

Here,  is the kinetic energy of the rocket and ${U_i}$ is the gravitational potential energy of the rocket.  

The kinetic energy of the rocket when it starts from the surface of Earth is:

${K_i} =\dfrac{1}{2}mv_i^2$

The total gravitational potential energy of the rocket when it was on the surface of the Earth is:

${U_i}= - \dfrac{{GMm}}{r}$

Here, $m$ is the mass of the rocket and $M$ is the mass of the Earth.

The final energy of the rocket will be in the form of the kinetic energy only. The final energy of the rocket will be:

${K_f} =\dfrac{1}{2}mv_f^2$

Now, using the conservation of energy for the rocket:

$\begin{aligned}T{E_i} &= T{E_f} \hfill\\{K_i} + {U_i} &= {K_f}\hfill\\\frac{1}{2}mv_i^2- \frac{{GMm}}{r}&= \frac{1}{2}mv_f^2 \hfill\\\end{aligned}$

Rearrange the above expression for final velocity and substitute the values.

$\begin{aligned}{v_f}&= \sqrt {v_i^2 - \frac{{GM}}{r}}\\&= \sqrt {{{\left( {1.80 \times {{10}^4}} \right)}^2} - \frac{{\left( {6.67 \times {{10}^{11}}} \right)\left( {5.97 \times {{10}^{24}}}\right)}}{{6.37 \times {{10}^6}}}}\\&= \sqrt {\left( {3.24 \times {{10}^8}} \right) - \left( {6.25 \times {{10}^7}} \right)} \\&= 1.61 \times {10^4}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

Thus, the speed of the rocket when it is very far away from the Earth is  $\boxed{1.61 \times {{10}^4}\,{{\text{m}}\mathord{\left/ {\vphantom{{\text{m}}{\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$

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Answer Details:

Grade: College

Subject: Physics

Chapter: Gravitation

Keywords:  Rocket, gravitational potential energy, kinetic energy, very far away, Earth, launched straight up, speed of rocket, initial energy, 1.80x10^4 m/s.

Answer 2

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:
$\frac{1}{2} mv_i ^2 - \frac{GM}{r} = \frac{1}{2}m v_f^2$
$v_f = \sqrt{v_i^2 - \frac{GM}{r} } =1.41 \cdot 10^4 m/s$