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2 years ago

7

a satellite is orbiting Earth at a distance of 35 kilometers. The satellite has a mass of 500 kilograms. what is the force betwe

en the planet and the satellite

2 answers:

ArbitrLikvidat [17]2 years ago

8 0

Answer:

The force between the planet and the satellite is 4.76 * 10³ Newtons

Explanation:

Given:

Mass of satellite = m = 500 kg

Distance of the satellite from Earth's surface = h = 35000 m

We know that:

Mass of Earth = M = 5.9 * 10²⁴ kg

Radius of Earth = R = 6.4 * 10⁶ m

Gravitational Constant = G = 6.673 x 10⁻¹¹ N m²/kg²

Force between Earth and an object is given as:

F = GmM/(R+h)²

= (6.673 x 10⁻¹¹ x 500 x 5.9 x 10²⁴)/((6.4 * 10⁶)+(3.5*10⁴))²

= (1.97*10¹⁷)/(6.435*10⁶)²

= (1.97*10¹⁷)/(4.14*10¹³)

= 4.76 * 10³ N

Naddik [55]2 years ago

4 0

Answer: F = 4.76 * 10³ N

Explanation:

We know that the force of gravity between two objects is:

F = G*m*M/r²

Where M is the mass of earth, m is the mass of the satelite, r is the distance between the radius of the earth and the satelite and G is the gravitational constant, and the data that we have is:

Mass of satellite: m = 500 kg

Distance of the satellite from Earth's surface: H = 0.035x10^6 m

Radius of Earth: R = 6.4 x 10⁶ m

So we have that r = R + H = 6.435x10^6

Mass of Earth: M = 5.9 * 10²⁴ kg

Gravitational Constant: G = 6.673 x 10⁻¹¹ N m²/kg²

Then the force that the Earth does in the satellite is:

F = (1.97*10¹⁷)/(6.435*10⁶)²

F = (1.97*10¹⁷)/(4.14*10¹³)

F = 4.76 * 10³ N

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A boy throws a water ballon such that it hits his sister standing 10m away. The boy threw the water balloon at an angle of 35 de

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Explanation:

We have the following data:

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On the other hand, when we talk about parabolic motion (as in this situation) the water ballon reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time $T$ it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

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Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (3)

Remembering $T=2t$ :

$T=2\frac{-V_{o}}{g}$ (4)

Substituting (4) in (1):

$x=V_{o}cos \theta (2\frac{-V_{o}}{g})$ (5)

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