Physics, 08.04.2020 00:15 weberalycia

A small asteroid that has a mass of 100 kg is moving at 200 m/s when it is 1000 km above the Moon. 1) How fast will the meteorite be traveling when it impacts the lunar surface if it is heading straight toward the center of the Moon

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Answer from: anannanekejdjd

The asteroid moves from a high elevation to a lower one therefore potential energy is converted to potential energy.

The speed of the meteorite when it impacts the lunar surface is approximately 1,449.9 m/s

Reasons:

The Kinetic and potential energies are;

Using the energy conservation equation, we get;

$-\dfrac{G \cdot M \cdot m}{R + h} + \dfrac{1}{2} \cdot m \cdot v_1^2 = -\dfrac{G \cdot M \cdot m}{R } + \dfrac{1}{2} \cdot m \cdot v_2^2$

Where;

G = Universal gravitation constant = 6.67408× 10⁻¹¹ m³/(kg·s²)

M = Mass of the Moon = 7.34767309 × 10²² kg

m = Mass of the asteroid = 100 kg

R = Radius of the Moon = 1,737.400 meters

Therefore, we get;

$\dfrac{G \cdot M \cdot m}{R } -\dfrac{G \cdot M \cdot m}{R + h} + \dfrac{1}{2} \cdot m \cdot v_1^2 = \dfrac{1}{2} \cdot m \cdot v_2^2$

$G \cdot M \cdot m\cdot \left(\dfrac{1}{R } -\dfrac{1}{R + h} \right)+ \dfrac{1}{2} \cdot m \cdot v_1^2 = \dfrac{1}{2} \cdot m \cdot v_2^2$

Dividing by m, gives;

$G \cdot M \cdot \left(\dfrac{1}{R } -\dfrac{1}{R + h} \right)+ \dfrac{1}{2} \cdot v_1^2 = \dfrac{1}{2} \cdot v_2^2$

Plugging in the values gives;

$6.67408 \times 10^{-11} \times 7.34767309 \times 10^ {22}\left(\dfrac{1}{1737400 } - \dfrac{1}{1737400 + 1 \times 10^6} \right)+ \dfrac{1}{2} \times 200^2 = \dfrac{1}{2} \cdot v_2^2$

Which gives;

$1051105.6191 = \dfrac{1}{2} \cdot v_2^2$

v₂² = 2× 1051105.6191 = 2102211.2382

v₂ = √(2102211.2382) ≈ 1,449.9

The speed of the meteorite when it impacts the lunar surface, v₂ ≈ 1,449.9 m/s

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Answer from: amuijakobp78deg

v = 200.005 m / s

Explanation:

For this exercise we can use the concept of energy conservation,

Starting point. When the asteroid at h = 1000 km = 1 10⁶ m

Em₀ = K + U = ½ m v₀² - G m M / (R + h)

Final point. When the asteroid is on the surface of the moon

Emf = K + U = ½ m v² - G m M / R

As there is no friction the energy is conserved

Em₀ = Emf

½ m v₀² - G m M / (R + h) = ½ m v² - G m M / R

½ m (v₀² –v²) = G m M (-1 / R + 1 / R + h)

½ (v₀² - v²) = G M (-h / R (R + h))

v² = v₀² + 2 G M h / R (R + h))

Let's calculate

v² = 200² + 2 6.67 10⁻¹¹ 7.36 10²² / 1.74 (1.74 + 1 ))10¹²

v2 = 40000 + 2,059

v = 200.005 m / s

Answer from: Quest

B) raising the temperature and reducing the volume of the gas.|-/

Answer from: Quest

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A small asteroid that has a mass of 100 kg is moving at 200 m/s when it is 1000 km above the Moon. 1...