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Elena-2011 [213]
2 years ago
15

Determine the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Eart

h is 1/60 of its value at the Earth's surface.
Physics
1 answer:
Temka [501]2 years ago
5 0

The distance from the Earth's center to the point outside the Earth is 55800 Km

<h3>How to determine the distance from the surface of the Earth</h3>
  • Acceleration due to gravity of Earth = 9.8 m/s²
  • Acceleration due to gravity of the poin (g) = 1/60 × 9.8 = 0.163 m/s²
  • Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
  • Mass of the Earth (M) = 5.97×10²⁴ Kg
  • Distance from the surface of the Earth (r) =?

g = GM / r²

Cross multiply

GM = gr²

Divide both sides by g

r² = GM / g

Take the square root of both sides

r = √(GM / g)

r = √[(6.67×10¯¹¹ × 5.97×10²⁴) / 0.163)]

r = 4.94×10⁷ m

Divide by 1000 to express in Km

r = 4.94×10⁷ / 1000

r = 4.94×10⁴ Km

<h3>How to determine the distance from the center of the Earth</h3>
  • Distance from the surface of the Earth (r) = 4.94×10⁴ Km
  • Radius of the Earth (R) = 6400 Km
  • Distance from the centre of the Earth =?

Distance from the centre of the Earth = R + r

Distance from the centre of the Earth = 6400 + 4.94×10⁴

Distance from the centre of the Earth = 55800 Km

Learn more about gravitational force:

brainly.com/question/21500344

#SPJ1

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Answer:

-24 m/s

Explanation:

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Also;

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Thus; the  initial velocity of the bowling ball when it first touched the mattress =  -24 m/s

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A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal ran
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Answer

a) For the rock

\dfrac{v_t^2sin 2\theta}{g} = \dfrac{v_t^2sin^2\theta}{2g}

2sin\thetacos\theta = \dfrac{sin^2\theta}{2}

2cos\theta = \dfrac{sin\theta}{2}

tan\theta = 4

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b) \theta = 45^0 for maximum range

\dfrac{d_{max}}{d}=\dfrac{(v_tcos 45^0)(2v_tsin 45^0)g}{(v_tcos 76^0)(2v_tsin 76^0)g}

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\dfrac{d_{max}}{d}=2.129

d_{max}=2.129 d

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<h3>What is the impact of speed on kinetic energy ?</h3>

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