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egoroff_w [7]
3 years ago
14

(a) Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and

place it in a circular low earth orbit, that is, an orbit whose altitude above the earth's surface is much less than RE. (As an example, the International Space Station is in low earth orbit at an altitude of about 400km, much less than RE = 6370km .) Ignore the kinetic energy that the spacecraft has on the ground due to the earth’s rotation.
(b) Calculate the minimum amount of additional work required to move the space craft from low earth orbit to a very great distance from the earth. Ignore the gravitational effects of the sun, the moon, and the other planets.
(c) Justify the statement "In terms of energy, low earth orbit is halfway to the edge of the universe.

Physics
1 answer:
notsponge [240]3 years ago
8 0

Answer:

(a) The Work need to   place spacecraft into low orbit isGmm_{E} /2R_{E}

(b) Additional Work need to  place the spacecraft far away from the earth is

Gmm_{E} /2R_{E}.

(c) From part(a) and part(b) we observe the Both quantities are equal.

Explanation:

Explanation is in the following attachments

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What phase does the moon have to be in for a solar eclipse
kakasveta [241]

Answer:

new moon

Explanation:

A solar eclipse take place at new moon phase, when the moon passes between the earth and the sun and its shadows fall on the Earth's surface which by definition a solar eclipse.

7 0
3 years ago
The temperature of the surface of the Sun is 5500°C.
natita [175]

Answer:

a. the average kinetic energy of hydrogen atoms is  1.20 × 10^-19J

b. the average kinetic energy of helium atoms is 1.24 × 10^-17J

Explanation:

The computation is shown below;

As we know that

Kinetic energy = 3 ÷ 2 kT

where,

K = Boltzmann constant

And, T = Temperature

a. Now the temperature in kelvin is

T = (5,500 × (°C ÷ K) + 273.15 K)

= 5773.15 K

As

Kinetic energy = 3 ÷ 2 kT

So now 1.38 × 10^-23 J/K for K would be substituted and 5773.15 K for Temperature T

Now Kinetic energy is

= 3 ÷ 2 (1.38 × 10^-23 J/K) ( 5773.15 K)

= 1.20 × 10^-19J

hence, the average kinetic energy of hydrogen atoms is  1.20 × 10^-19J

b. As

Kinetic energy = 3 ÷ 2 kT

now 1.38 × 10^-23 J/K for K would be substituted and 6 × 10^5K for Temperature T

Now Kinetic energy is

= 3 ÷ 2 (1.38 × 10^-23 J/K) (6 × 10^5K )

= 1.24 × 10^-17J

hence, the average kinetic energy of helium atoms is 1.24 × 10^-17J

8 0
3 years ago
is a climate cycle where the temperature of the Ocean changes because of movements of air and ocean currents. During this period
k0ka [10]
<h2>Answer: <u>El Niño</u> phenomenon </h2>

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To understand it better, it is necessary to know the following:

Normally in the referred area of the Pacific, trade winds blow from east to west, which move the warm water from the area to the west, allowing deeper and colder water to emerge. But, during El Niño the trade winds weaken or even flow in reverse, from west to east. It is then, when the warm water accumulates in front of the pacific coasts of South America provoking a change in the habitual patterns of precipitation and temperature.

As a consequence, hot water evaporates and condenses, causing an increase in rainfall. Due to this change in the trade winds, the climate pattern is altered, and these rains affect more the eastern part of the Pacific, while at the other end of the ocean (in Australia and Southeast Asia) the climate becomes colder and dry.

It should be noted that this is a global phenomenon that affects and extends directly or indirectly in most regions of the planet. However, its greatest impact occurs in the American countries of the Pacific coast; and in Southeast Asia and Australia.

This phenomenom begins during the final months of the year, from October and generally may last until January of the following year, although its effects are often maintained even until March.

3 0
3 years ago
A raindrop has a mass of 7.7 × 10-7 kg and is falling near the surface of the earth. Calculate the magnitude of the gravitationa
goldfiish [28.3K]

Given that the mass of the raindrop is

m=7.7\times10^{-7}\text{ kg}

The acceleration due to gravity is g = 9.81 m/s^2

We have to find

(a) Magnitude of the gravitational force exerted on the raindrop by earth

(b)Magnitude of the gravitational force exerted on the earth by the raindrop

(a) The formula to calculate the magnitude of the gravitational force exerted on the raindrop by the earth is

F=mg

Substituting the values, the gravitational force exerted on the raindrop by the earth is

\begin{gathered} F=7.7\times10^{-7}\times9.81 \\ =7.55\times10^{-6}\text{ N} \end{gathered}

(b) According to Newton's third law,

If F' is the gravitational force exerted on the earth by the raindrop, then

F=-F^{\prime}

Here, the negative sign indicates that both forces act in opposite direction.

The gravitational force exerted on the earth by the raindrop is

F^{\prime}\text{ = -7.55}\times10^{-6}\text{ N}

And the magnitude of the gravitational force exerted on the earth by the raindrop is

7.55\times10^{-6}\text{ N}

Thus, the magnitude of both forces is equal.

5 0
1 year ago
Is it possible for the velocity of an object to be zero and its acceleration not zero?
Nesterboy [21]

Explanation:

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8 0
3 years ago
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