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RUDIKE [14]
2 years ago
15

Calculate the gravitational force of the Earth and Moon. The Earth has a mass of 5.972x 1024 kg and the Moon has a mass of 7.348

x 1022 kg. They are an average of 384 million meters apart.
Physics
2 answers:
Wewaii [24]2 years ago
7 0

The gravitational force is 1.98×10^20 m.

Computation of gravitational force:

The gravitational force F between two masses is given by the formula,

F=Gm1m2/ r^2

where G is the gravitational constant.

Note: It is assumed m1=5.972×10^24 kg, m2=7.348×10^22 kg and r=3.84×10^8 m and G=6.67×10^(-11) N m^2 kg^-2

Then F=(6.67×10^(-11)×5.972×10^(24)×7.348×10^(22))/ (3.84×10^8)

         F=1.98×10^(20) N

Check out more about gravitational force here:

brainly.com/question/13054973

#SPJ 4

Len [333]2 years ago
6 0

Answer:

The force of gravitation is 1.98×10^(20) m.

Explanation:

Note: The value of m1, m2, r, and G is taken to be 5.97×10^(24) kg, 7.348×10^(22) kg, 3.84×10^(8) m, and 6.67×10^(-11) N m^2/ kg^2 respectively.

The force of gravitation F acting between two masses is calculated using the formula,

F=Gm1m2/ r^2

where G is the gravitational constant.

Hence force F is

F=6.67×10^(-11)×5.97×10^(24)×7.348×10^(22)/ (3.84×10^(8))^2

F=1.98×10^(20) N

Check out other solutions.

brainly.com/question/13054973

#SPJ10

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What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 203 km above Earth's surface
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Answer:

7,790.38 m/s

Explanation:

Given the following :

What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 203 km above Earth's surface

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Using the formula :

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Where G = gravitational constant =6.67×10^-11

Kindly check attached picture for detailed explanation.

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You wish to place a spacecraft in a circular orbit around the earth so that its orbital speed will be 4.00×103m/s. What is this
Levart [38]

Answer:

The radius of orbit=2.49\times 10^7 m

Explanation:

We are given that

Orbital speed=4.00\times 10^3m/s

We have to find the radius of orbit of spacecraft.

We know that

Gravitational constant=6.67\times 10^{-11}m^3/kgs^2

Mass of earth=5.972\times 10^{24} kg

Orbital speed=\sqrt{\frac{GM}{r}}

Where G= Gravitational constant

M=Mass of earth

r=Radius of orbit

Substitute the values in the formula

4\times 10^3=\sqrt{\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{r}

Squaring on both sides

16\times 10^6=\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{r}

r=\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{16\times 10^6}

r=2.49\times 10^7 m

Hence, the radius of orbit=2.49\times 10^7 m

7 0
4 years ago
A child pulls a wagon at a constant velocity of 4.0 m/s for 4.0 minutes along a level sidewalk. The child does this applying a 2
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Explanation:

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You are a geologist studying the electrical field on a given region. You find that the electric field is vertical and perpendicu
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Answer:

charge      Qint = 7.17 10⁻⁴ C

Explanation:

For this problem we must use Gauss's law

       F = ∫ E. dA = Qint / εₙ

let's form a Gaussian surface that is parallel to the surface, for example, a Cube. As the field is vertical and perpendicular to the surface, the field lines and the area vector are parallel whereby the scalar product is reduced to an ordinary product.

    Φ = E A = Qint / ε₀

   

    A = 1 km² (1000 m / 1km)² = 1 10⁶ m²

We can calculate the charge

    Qint = E A ε₀

    Qint = 81 1 10⁶ 8.85 10⁻¹²

    Qint = 7.17 10⁻⁴ C

5 0
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