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jek_recluse [69]
3 years ago
15

In 1610, galileo used his telescope to discover four prominent moons around jupiter. their mean orbital radii a and periods t ar

e as follows: (a) io has a mean orbital radius of 4.22 x 108 m and a period of 1.77 days. find the mass of jupiter from this information. (b) europa has a mean orbital radius of 6.71 x 108 m and a period of 3.55 days. find the mass of jupiter from this information. (c) ganymede has a mean orbital radius of 10.7 x 108 m and a period of 7.16 days. find the mass of jupiter from this information. (d) callisto has a mean orbital radius of 18.8 x 108 m and a period of 16.7 days. find the mass of jupiter from this information
Physics
1 answer:
katrin2010 [14]3 years ago
6 0

Time period of any moon of Jupiter is given by

T = 2\pi \sqrt{\frac{r^3}{GM}}

from above formula we can say that mass of Jupiter is given by

M = \frac{4 \pi^2 r^3}{GT^2}

now for part a)

r = 4.22 * 10^8 m

T = 1.77 day = 152928 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (4.22 * 10^8)^3}{(6.67 * 10^{-11})(152928)^2}

M = 1.9* 10^{27} kg

Part B)

r = 6.71 * 10^8 m

T = 3.55 day = 306720 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (6.71 * 10^8)^3}{(6.67 * 10^{-11})(306720)^2}

M = 1.9* 10^{27} kg

Part c)

r = 10.7 * 10^8 m

T = 7.16 day = 618624 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (10.7 * 10^8)^3}{(6.67 * 10^{-11})(618624)^2}

M = 1.89* 10^{27} kg

PART D)

r = 18.8 * 10^8 m

T = 16.7 day = 1442880 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (18.8 * 10^8)^3}{(6.67 * 10^{-11})(1442880)^2}

M = 1.889* 10^{27} kg

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ON
mojhsa [17]

B. the distance the star is from Earth

Explanation:

The apparent magnitude of star is a function of its distance from the earth. It is one of the physical properties that is used to study a star.

The apparent magnitude of a star or other astronomical bodies is a measure of their brightness as seen from a location on the earth.

The apparent magnitude of a star depends on:

  • Distance of the star from the location on earth.
  • luminosity of the star
  • the particles along the part of the star and earth that cuts off the light the earth receives.

learn more:

Star luminosity brainly.com/question/9084808

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5 0
4 years ago
A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.
EleoNora [17]

(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

(b) The work done  in firing the projectile is 2,500 J.

<h3>Kinetic energy of the projectile at maximum height</h3>

The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;

K.E = ¹/₂mv₀ₓ²

where;

  • m is mass of the projectile
  • v₀ₓ is the initial horizontal component of the velocity at maximum height

<u>Note:</u> At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.

K.E = (0.5)(2)(30²)

K.E = 900 J

<h3>Work done in firing the projectile</h3>

Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.

W = K.E(i) = ¹/₂mv²

where;

  • v is the resultant velocity

v = √(30² + 40²)

v = 50 m/s

W = (0.5)(2)(50²)

W = 2,500 J

Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

The work done  in firing the projectile is 2,500 J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

5 0
2 years ago
A beam of red light is made to pass through two slits that are 3.55 E-3 meters apart. On a screen 2.25 meters away from the slit
JulsSmile [24]
I am assuming you know the relation obtained between slit width, distance of screen from slits, distance of interference pattern obtained on the screen from the center and the wavelength of monochromatic light used in Young's Double Slit experiment.
λ =\frac{y*d}{D} =  \frac{3.55*10^{-3}*1.25*10^{-4}  }{2.25} = 1.97*10^{-7} m
λ ~ 1.97 ×10⁻⁷m
7 0
3 years ago
The atmosphere protects us from all of the following except Meteorites The sun's rays Pollution Wind
denis23 [38]
Meteorites is the answer

6 0
3 years ago
HELP ASAP!!! The gravitational pull of the Moon is much less than the
katovenus [111]

Answer:

I'm pretty sure its B and C

Explanation:

B bc the weight is gravitational pull x mass so when the object has same mass the weight is smaller on moon

C bc mass is the same - you can't change it

7 0
3 years ago
Read 2 more answers
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