Question

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be when it is very far from the Earth

Answer 1

Answer:

The value is  $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

  The  initial speed is $u = 2.05 *10^{4} \ m/s$

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             $T__{E}} = KE__{i}} + KE__{e}}$

Here  $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          $KE_i = \frac{1}{2} * m * u^2$

=>       $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=>       $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And  $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       $KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=>    $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=>    $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        $KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

        $T__{E}} = KE_p$

So

       $2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=>     $5.4564 *10^{8} = v^2$

=>     $v = \sqrt{5.4564 *10^{8}}$

=>     $v = 2.3359 *10^{4} \ m/s$