Physics, 30.07.2019 21:30 Shaylaaaaaaaa

Two meteoroids are heading for earth. their speeds as they cross the moon's orbit are 1.0 km/s . part a the first meteoroid is heading straight for earth. what is its speed of impact?

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Answer from: donaji1024perez

Answer : $v_{2} = 11.2\ km/s$

Explanation : The total energy of the meteoroids

Total Energy = Kinetic Energy + Potential Energy

$E = \dfrac{1}{2}mv^{2} - \dfrac{GMm}{R}$

Total energy remain constant.

Now, for both meteoroids

$\dfrac{1}{2}m(v_{2}^{2} - v_{1}^{2}) = GMm\left(\dfrac{1}{R}-\dfrac{1}{r}\right)$

Where, R = radius of the earth

r = distance of the moon

M = mass of the earth

m = mass of the meteoroid

$v_{2}^{2} = 2GM\left(\dfrac{1}{R}-\dfrac{1}{r}\right)+ v_{1}^{2}.....(I)$

Now, put the values in equation (I)

$v_{2}^{2} = 1 + 2\times6.67\times10^{-11}\times6\times10^{24}\left(\dfrac{1}{6.37} - \dfrac{1}{384}\right)\times10^{-6}$

$v_{2}^{2} = 12361\times 10^{4}\ m/s$

$v_{2} = 11118.003\ m/s$

$v_{2} = 11.2\ km/s$

Hence, this is the required solution.

Answer from: cruzhazeL999

The speed of the meteorite at its impact on Earth will be $\boxed{11.1\,{{{\text{km}}}\mathord{\left/{\vphantom{{{\text{km}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ or $\boxed{11088.24\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ .

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:

${E_k} = \frac{1}{2}m{v^2}$

The potential energy of the meteorite at a particular height from Earth is given by:

${E_p} = - \frac{{GMm}}{r}$

Here, is the mass of the Earth, is the mass of satellite and is the distance of the meteorite from the centre of the Earth.

The total energy of the meteorite is given as:

$\begin{aligned}{E_T}&={E_k} + {E_p}\\&=\frac{1}{2}m{v^2}-\frac{{GMm}}{r}\\\end{aligned}$

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:

$\begin{aligned}\frac{1}{2}mv_1^2-\frac{{GMm}}{r}&=\frac{1}{2}mv_2^2-\frac{{GMm}}{R} \hfill\\v_1^2-v_2^2&=2GM\left({\frac{1}{r}-\frac{1}{R}}\right) \hfill\\{v_2}&=\sqrt {v_1^2-2GM\left({\frac{1}{r}-\frac{1}{R}}\right)}\hfill\\\end{aligned}$

Here, ${v_1}$ is the speed of the meteorite in moon’s orbit, is the distance of the moon’s orbit from centre of Earth and is the radius of Earth.

The meteorite passes the moon’s orbit at the speed of $1\,{{{\text{km}}}\mathord{\left/{\vphantom{{{\text{km}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ .

We know that the distance of the Moon’s orbit from the centre of Earth is $3.84 \times{10^5}\,{\text{km}}$ and the radius of the Earth is $6371\,{\text{km}}$ .

Substitute the values in the equation of Energy.

$\begin{aligned}v_{f}&=\sqrt{(1)^{2}-2(6.67\times10^{-11})(5.97\times10^{24})\left(\dfrac{1}{384}-\dfrac{1}{6.37}\right)10^{-6}}\\&=\sqrt{1.229\times10^{8}}\text{ m/s}\\&=11088.24\text{ m/s}\\&\approx11.1\text{ km/s}\end{aligned}$

Thus, the speed of the meteorite striking the Earth is $\boxed{11.1\,{{{\text{km}}}\mathord{\left/{\vphantom{{{\text{km}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ or $\boxed{11088.24\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ .

Learn More:

1. Calculate the total force on the earth due to Venus, Jupiter, and Saturn

2. Compare the surface area–to–volume ratios of Earth and Venus

3. A rocket being thrust upward as the force of the fuel being burned

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.

Answer from: Quest

i think the answer is a and b

i wish you luck

Answer from: Quest

part 1: which substance has a higher solubility?

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by

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