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Stells [14]
2 years ago
5

(a) Determine the required delta-v, Ave, to the nearest m/s, to reach a circular 500 km altitude equatorial prograde (eastward)

orbit from a launch site on the Equator of the Earth. Assume the required velocity change is applied instantaneously at launch and neglect the effects of drag and any perturbation forces other than central gravity (b) If one instead wants to launch from the same Equatorial site into a circular 500 km altitude polar orbit, determine the required delta-v, Avp, to the nearest m/s.
Physics
1 answer:
yaroslaw [1]2 years ago
5 0

Answer:

a)    v₁ = 8.20 10³ m / s , b)     v₁’= 8.4 10³ m / s

Explanation:

a) For this exercise we can use energy conservation, in two points

Initial. On the earth's surface

             Em₀ = K + U = ½ m v₁² - G m Me / Re²

Final. In the orbit at a height of h = 500 10³m

             Em_{f} = K + U = ½ m v₂² - G m Me / (Re+h)²

            Em₀ = Em_{f}

            ½ m v₁² - G m Me / Re² = ½ m v₂² - G m Me / (Re+h)²

           ½ v₁² - ½ v₂² = G Me (1 / Re² - 1 / (Re + h)²)

To find the speed of the rocket in orbit we use Newton's sunga law, where force is the force of gravitation

              F = m a

The acceleration is centripetal

            a = v² / r

            G m Me / (Re+h) 2 = m v2 / (Re+h)

            v₂² = G Me / (Re + h)

  Let's calculate

             v² = √ (6.67 10⁻¹¹ 5.98 10²⁴ / (6.37 10⁶ +0.5 10⁶)

             v² = √ 58.05909 10⁶

             v₂ = 7.62 10³ m / s

Now we can find the clearance speed (v1)

             v₁² = 2 G Me (1 / Re2 - 1 / (Re + h) 2) + v₂²

             

Let's calculate

            v₁² = 2 6.67 10⁻¹¹ 5.98 10²⁴ (1 / 6.37 10⁶ - 1 / 6.87 10⁶) + 58.059 10⁶

            v₁² = 0.91145 10⁷ + 5.8059 10⁷

            v₁ = √ 67.172 10⁶

            v₁ = 8.20 10³ m / s

b) If the same place is launched into a polar orbit, the fundamental change is the distance that we can enter using the Pythagorean theorem.

               

              h’= √ 500 + 5002 = 500 √2

              h’= √ 2 0.5 10⁶ m

              Re + h’= 6.37 10⁶ + √2 0.5 10⁶

               Re + h’= 7.077 10⁶ m

Calculate

             v₁²’= 2 6.67 10⁻¹¹ 5.98 10²⁴ (1 / 6.37 10⁶ - 1 / 7.077 10⁶) + 58.059 10⁶

             v₁²’= 1,251 10⁷ + 5,8059 10⁷

             v₁’= 8.4 10³ m / s

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lyudmila [28]

Answer:

100m \div  \frac{9 \: m}{1sec}  = 11.11 \: sec \:

I guess you can round it to 11 seconds.

Explanation:

Going with a speed 9m/s means you are going 9 meters in each second.

If you are going 9 meters in second how many seconds will it take to 100 meters?

Visually;

9 meters - - - 1 second

100 meters - - - ?seconds.

When you write like this 9 times ?seconds equal to 100 meters time 1 second. (you probably know this but just in case)

So to find ?second you multiply 100meters by 1 and divide it by 9 whixh will give you 11.1111 seconds whixh again I believe you can round it to 11.

(Kind of a) Proof;

If 9m * ?sec = 100 m * 1 sec

you send 9 meters to other side.

?sec = (100 m * 1 sec) ÷ 9m

Hope it was clear and it helps! Please let me know if you have any questions.

7 0
3 years ago
A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H
Snowcat [4.5K]

Explanation:

Given

initial velocity(v_0)=1.72 m/s

\theta =44.8{\circ}

using v^2-u^2=2as

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration   (gsin\theta )

s=distance traveled

0-(1.72)^2=2(-9.81\times sin(44.8))s

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

0=1.72-gsin(44.8)\times t

t=\frac{1.72}{9.8\times sin(44.8)}

t=0.249 s

(c)Speed of the block at bottom

v^2-u^2=2as

Here u=0 as it started coming downward

v^2=2\times gsin(44.8)\times 0.214

v=\sqrt{2.985}

v=1.72 m/s

3 0
3 years ago
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6 0
3 years ago
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When electromagnetic fields interact with charged particles,
jasenka [17]
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5 0
3 years ago
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Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram
Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

  • Force 1 = 500 Newton
  • Distance 1 = 1.2 meter
  • Force 2 = 500 Newton
  • Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

Work\;done = Force \times distance

<u>For </u><u>student 1</u><u>:</u>

Work\;done = 500 \times 1.2

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

Work\;done = 500 \times 5

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

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2 years ago
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