Respuesta :
Answer : [tex]v_{2} = 11.2\ km/s[/tex]
Explanation : The total energy of the meteoroids
Total Energy = Kinetic Energy + Potential Energy
[tex]E = \dfrac{1}{2}mv^{2} - \dfrac{GMm}{R}[/tex]
Total energy remain constant.
Now, for both meteoroids
[tex]\dfrac{1}{2}m(v_{2}^{2} - v_{1}^{2}) = GMm\left(\dfrac{1}{R}-\dfrac{1}{r}\right)[/tex]
Where, R = radius of the earth
r = distance of the moon
M = mass of the earth
m = mass of the meteoroid
[tex]v_{2}^{2} = 2GM\left(\dfrac{1}{R}-\dfrac{1}{r}\right)+ v_{1}^{2}.....(I)[/tex]
Now, put the values in equation (I)
[tex]v_{2}^{2} = 1 + 2\times6.67\times10^{-11}\times6\times10^{24}\left(\dfrac{1}{6.37} - \dfrac{1}{384}\right)\times10^{-6}[/tex]
[tex]v_{2}^{2} = 12361\times 10^{4}\ m/s[/tex]
[tex]v_{2} = 11118.003\ m/s[/tex]
[tex]v_{2} = 11.2\ km/s[/tex]
Hence, this is the required solution.
The speed of the meteorite at its impact on Earth will be [tex]\boxed{11.1\,{{{\text{km}}}\mathord{\left/{\vphantom{{{\text{km}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] or [tex]\boxed{11088.24\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] .
Further Explanation:
The meteorite in the outer space at a particular distance from the Earth has its energy stored in the form of the kinetic energy as well as the potential energy.
The kinetic energy of the meteorite is given by:
[tex]{E_k} = \frac{1}{2}m{v^2}[/tex]
The potential energy of the meteorite at a particular height from Earth is given by:
[tex]{E_p} = - \frac{{GMm}}{r}[/tex]
Here, [tex]M[/tex] is the mass of the Earth, [tex]m[/tex] is the mass of satellite and [tex]r[/tex] is the distance of the meteorite from the centre of the Earth.
The total energy of the meteorite is given as:
[tex]\begin{aligned}{E_T}&={E_k} + {E_p}\\&=\frac{1}{2}m{v^2}-\frac{{GMm}}{r}\\\end{aligned}[/tex]
As the meteorite moves from the moon’s orbit towards the Earth, the potential energy of the meteorite gets converted into the kinetic energy.
Using conservation of energy for the meteorite from the moon’s orbit to the Earth’s surface:
[tex]\begin{aligned}\frac{1}{2}mv_1^2-\frac{{GMm}}{r}&=\frac{1}{2}mv_2^2-\frac{{GMm}}{R} \hfill\\v_1^2-v_2^2&=2GM\left({\frac{1}{r}-\frac{1}{R}}\right) \hfill\\{v_2}&=\sqrt {v_1^2-2GM\left({\frac{1}{r}-\frac{1}{R}}\right)}\hfill\\\end{aligned}[/tex]
Here, [tex]{v_1}[/tex] is the speed of the meteorite in moon’s orbit, [tex]r[/tex] is the distance of the moon’s orbit from centre of Earth and [tex]R[/tex] is the radius of Earth.
The meteorite passes the moon’s orbit at the speed of [tex]1\,{{{\text{km}}}\mathord{\left/{\vphantom{{{\text{km}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] .
We know that the distance of the Moon’s orbit from the centre of Earth is [tex]3.84 \times{10^5}\,{\text{km}}[/tex] and the radius of the Earth is [tex]6371\,{\text{km}}[/tex] .
Substitute the values in the equation of Energy.
[tex]\begin{aligned}v_{f}&=\sqrt{(1)^{2}-2(6.67\times10^{-11})(5.97\times10^{24})\left(\dfrac{1}{384}-\dfrac{1}{6.37}\right)10^{-6}}\\&=\sqrt{1.229\times10^{8}}\text{ m/s}\\&=11088.24\text{ m/s}\\&\approx11.1\text{ km/s}\end{aligned}[/tex]
Thus, the speed of the meteorite striking the Earth is [tex]\boxed{11.1\,{{{\text{km}}}\mathord{\left/{\vphantom{{{\text{km}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] or [tex]\boxed{11088.24\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] .
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Answer Details:
Grade: College
Subject: Physics
Chapter: Gravitation
Keywords:
Meteorite, moon’s orbit, 1km/s, two meteorite, heading for, speed of impact, 11.1 km/s, 384000 km, total energy, heading straight.
