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Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

a

The orbital speed is $v= 2.6*10^{3} m/s$

b

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

2 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!<br> I NEED THE ANSWER NOW

FrozenT [24]

90 F = 43 OR 0.9F = 0.43
(F = 43 / 90 OR 0.43 / 0.9 =) 0.48 N

upwards force = downwards force
(R =) 1.2 N

5 0

2 years ago

in an isolated system, two cars, each with a mass of 1,500 kg, collide. car 1 is initially at rest while car 2 is moving at 5 m/

kirza4 [7]

Answer:2.5 m/s

Explanation:

6 0

2 years ago

Complete the passage to describe the relationship between intensity and loudness. Intensity is the amount of ______ that flows i

Usimov [2.4K]

Answer:

Blank One: I'm not certain but I'd assume it's either sound or energy.

Blank Two: Louder

Explanation:

volume is controlled by how much energy is put in an area per a unit of time, so the more energy in one area for a unit of time, the louder the sound will come out.

Sorry I couldn't give the best explanation!

7 0

2 years ago

Read 2 more answers

In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha

Assoli18 [71]

Answer:

$\mathbf{v_a = 4.06 \times 10^7 \ m/s}$

Explanation:

$\text{The missing part of the question is attached below.} \\ \\ \text{ mass m = 4u and charge q = 2e. Suppose a uranium nucleus with 92 protons decays into }$ $\text{into thorium, with 90 protons, and an alpha particle. The alpha particle is initally at rest at }$ $\text{ the surface of the thorium nucleus, which is 15 fm in diameter. What is the speed of the alpha}$ $\text{particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest. }$

$\text{So, from the above infromation;}$

$radius (r) = \dfrac{15 \ fm }{2}$

$radius (r) = 7.5 \times 10^{-15} \ m$

$\text{Using the formula for potential energy}$

$U = \dfrac{(9\times 10^9 \ Nm^2/C^2)(2(1.6\times10^{-19} \ C ))(90)(1.6\times 10^{-19} C)}{7.5 \times 10^{-13} \ m}$

$U = 5.529 \times 10^{-12} \ J$

$\text{Now, the speed of the alpha particle can be estimated from the conservation of energy principle}$ $\dfrac{1}{2}mv_a^2= 5.529 \times 10^{-12} J$

$v_a = \sqrt{\dfrac{2(5.529 \times 10^{-12} \ J)}{4(1.67 \times 106{-37} \ kg)}}$

$\mathbf{v_a = 4.06 \times 10^7 \ m/s}$

4 0

2 years ago