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3 years ago

10

An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.25 m/s2.

Determine the orbital period of the satellite.

2 answers:

Rama09 [41]3 years ago

8 0

Answer:

T = 7101 s = 118.35 mins = 1.9725 hrs

Explanation:

To solve the question, we apply the formula for gravitational acceleration

a = GM/r², where

a = acceleration due to gravity

G = gravitational constant

M = mass of the earth

r = distance between the satellite and center of the earth

Now, if we make r, subject of formula, we have

r = √(GM/a)

Recall also, that

a = v²/r, making v subject of formula

v = √ar

If we substitute the equation of r into it, we have

v =√a * √r

v =√a * √[√(GM/a)]

v = (GM/a)^¼

Again, remember that period,

T = 2πr/v, we already have v and r, allow have to do is substitute them in

T = 2π * √(GM/a) * [1 / (GM/a)^¼]

T = 2π * (GM/a³)^¼

T = 2 * 3.142 * [(6.67*10^-11 * 5.97*10^24) / (6.25³)]^¼

T = 6.284 * [(3.982*10^14) / 244.140]^¼

T = 6.284 * (1.63*10^12)^¼

T = 6.284 * 1130

T = 7101 s

T = 118.35 mins

T = 1.9725 hrs

s344n2d4d5 [400]3 years ago

5 0

Answer:

118 minutes( 2 hours approximately )

Explanation:

Here, we are interested in calculating the orbital period of the satellite

Please check attachment for complete solution

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A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is

lesantik [10]

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth = 5.98 × 10²⁴ kg

Radius of Earth = $6.38\times 10^6\ m$

The gravitational force will balance the centripetal force

$\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}$

$T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}$

From Kepler's law we have relation

$T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m$

Distance from the center of the Earth would be

$42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}$

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aniked [119]

Complete Question:

Each lunar cycle has one full moon, in which the relative positions of Earth, the sun, and the moon form in a straight line. Which list represents the position of Earth, the sun, and the moon during a full moon?

Group of answer choices.

A. Earth, sun, moon

B. sun, moon, Earth

C. moon, sun, Earth

D. sun, Earth, moon

Answer:

D. sun, Earth, moon

Explanation:

A lunar eclipse is a phenomenon that occurs when the Earth comes between the Moon and the Sun thereby causing it to cover the Moon with its shadow.

Simply stated, lunar eclipse takes place when the Moon passes or moves through the Earth's shadow thereby blocking any ray of sunlight from reaching the Moon. Thus, the full moon appears deep red (blood moon).

Also, a lunar eclipse would occur only when the Sun, Earth, and Moon are closely aligned to form a straight line known as the syzygy.

There are three (3) types of lunar eclipse and these are;

1. Total lunar eclipse.

2. Partial lunar eclipse.

3. Penumbra lunar eclipse.

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3 0

3 years ago

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

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$\tau=Fd$

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

$\tau=(25 N)(0.5 m)=12.5 Nm$

4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity ( $\omega$ ) is:

$L=I\omega$

In this problem, we have

$I=0.007875 kgm^2$

$\omega=6.28 rad/s$

So, the angular momentum is

$L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s$

6 0

3 years ago