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worty [1.4K]
3 years ago
13

Newton’s law of universal gravitation.. . a.. is equivalent to Kepler’s first law of planetary motion.. b.. can be used to deriv

e Kepler’s third law of planetary motion.. c.. can be used to disprove Kepler’s laws of planetary motion.. d.. does not apply to Kepler’s laws of planetary motion.
Physics
2 answers:
mixer [17]3 years ago
8 0
By the help of newtons law of gravitation we can derive keplers third law of planetary motion.
emmainna [20.7K]3 years ago
6 0

Answer:

b.. can be used to derive Kepler’s third law of planetary motion..

Explanation:

As per Newton's law of gravitation we can say that two masses separated by some distance "r" will attract each other with force given by

F = \frac{Gm_1m_2}{r^2}

now we can say that

if all planets revolves around the sun in approx circular orbit then the centripetal force on the planets will be due to gravitational force of Sun

so we can say

\frac{mv^2}{R} = \frac{GMm}{R^2}

so by solving above equation we can find average orbital speed of all planets and hence we can derive all Kepler's law

so correct answer will be

b.. can be used to derive Kepler’s third law of planetary motion..

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Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks
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The concept of this question can be well understood by listing out the parameters given.

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The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

\mathsf{S = ut + \dfrac{1}{2}gt^2}

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

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The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}

\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}

\mathsf{0.3 =4.9*(t^2)}

By dividing both sides by 4.9, we have:

\mathsf{t^2 = \dfrac{0.3}{4.9}}

\mathsf{t^2 = 0.0612}

\mathsf{t = \sqrt{0.0612}}

\mathsf{t =0.247  \ seconds}

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}

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\mathsf{t = \dfrac{T}{2}}

T = 2t

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T = 0.494 seconds

\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}

By making the spring constant k the subject of the formula:

\mathsf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}

\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}

\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}

\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\  \mathsf{  k = \dfrac{2 \pi^2*m}{T^2}}

\mathsf{  k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}

\mathbf{  k =8.25 \ N/m}

Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

brainly.com/question/17315536?referrer=searchResults

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