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3 years ago

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Two planets orbit a far away star as shown. Is it possible that both planets experience the same gravitation force from the star

?
A) No, it is not possible.
B) Yes, it is possible if the closer planet has less mass.
C) Yes, it is possible if the closer planet as a smaller radius.
D) Yes, it is possible if the closer planet orbits the star in a shorter time.

1 answer:

Vesna [10]3 years ago

8 0

As we know that gravitational force on two planets will be given as

let

mass of star = M

mass of two planets are m1 and m2

their distance from star is r1 and r2

$F = \frac{GMm_1}{r_1^2} = \frac{GMm_2}{r_2^2}$

since the gravitational force of star is given on two planets to be same

so here we can say

$\frac{m_1}{r_1^2} = \frac{m_2}{r_2^2}$

now if the ratio of mass and distance is same then we can say that closer planet must have lesser mass to hold good for above equation

so correct answer will be

B) Yes, it is possible if the closer planet has less mass.

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Answers:

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Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum $p_{i}$ before the collision must be equal to the final momentum $p_{f}$ after the collision:

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Being:

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$p_{f}=(M+m) V$

Where:

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$V_{i}=45 m/s$ the initial speed of the falcon

$m=240 g \frac{1 kg}{1000 g}=0.24 kg$ is the mass of the pigeon

$U_{i}=0 m/s$ the initial speed of the pigeon (at rest)

$V$ the final speed of the system falcon-pigeon

Then:

$MV_{i}+mU_{i}=(M+m) V$ (2)

Finding $V$ :

$V=\frac{MV_{i}}{M+m}$ (3)

$V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}$ (4)

$V=30 m/s$ (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

$F=\frac{\Delta p}{\Delta t}$ (6)

Where:

$F$ is the force exerted on the pigeon

$\Delta t=0.015 s$ is the time

$\Delta p$ is the pigeon's change in momentum

Then:

$\Delta p=p_{f}-p_{i}=mV-mU_{i}$ (7)

$\Delta p=mV$ (8) Since $U_{i}=0$

Substituting (8) in (6):

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$F=\frac{(0.24 kg)(30 m/s)}{0.015 s}$ (10)

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Answer:

Explanation:

Given that,

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Force constant K = 8N/m

Amplitude A = 11.6 cm

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Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

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Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

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c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

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9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

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t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

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d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

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