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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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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

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Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

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Answer:

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Specifically the Sun has a color of 550 nm which corresponds to 5400K

bright stars have a BLUE color corresponding to 7500K

the brightest found are Blue - White with a temperature of 20000K

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