1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
artcher [175]
2 years ago
8

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

has been found to have 2.3 times the earth's diameter and 7.9 times the earth's mass. Observations of this planet over time show that it is in a nearly circular orbit around its star and completes one orbit in only 9.5 days. How many times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun. Assume that the earth is also in a nearly circular orbit.
a.) 0.026r
b.) 0.078r
c.) 0.70r
d,) 2.3r
Physics
1 answer:
Stels [109]2 years ago
7 0

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

               \frac{G M m}{r^{2}}=m \omega^{2} r

The orbit’s period is given by,

               T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}

Where,

T_{e} = Earth’s period

T_{p} = planet’s period

M_{s} = sun’s mass

r_{e} = earth’s radius

Now,

             T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}

As, planet mass is equal to 0.7 times the sun mass, so

            T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}

Taking the ratios of both equation, we get,

             \frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}

            \frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}

           \frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}

Given T_{p}=9.5 \text { days } and T_{e}=365 \text { days }

          \frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}

         r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}

You might be interested in
Pleaseeeeeeeeee helpppppppppppppppp
Tju [1.3M]
I feel like it could be A
8 0
2 years ago
Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal ener
AVprozaik [17]

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

=  1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

8 0
2 years ago
Calculate the percentage of an iceberg submerged beneath the surface of the ocean given that the density of ice is 916.3kg/m3 an
pogonyaev

Answer:

The percentage of an iceberg submerged beneath the surface of the ocean = 89.67%

Explanation:

Let V be the total volume of the iceberg

Let x be the volume of iceberg submerged

According to Archimedes principle,

weight of the iceberg = weight of the water displaced (that is, weight of x volume of water)

Weight of the iceberg = mg= ρ(iceberg) × V × g

Weight of water displaced = ρ(fluid) × x × g

We then have

ρ(iceberg) × V × g = ρ(fluid) × x × g

(x/V) = ρ(iceberg) ÷ ρ(fluid) = 916.3 ÷ 1021.9 = 0.8967 = 89.67%

Hope this Helps!!!!

6 0
2 years ago
Which of the following statements can be inferred given the food web shown above? A. Energy flows from consumers to producers wi
Zarrin [17]

Answer:A.

Explanation:

4 0
2 years ago
Read 2 more answers
A single point on a distance time graph tells the
Sonbull [250]

Answer: Instantaneous speed.

Explanation:

4 0
3 years ago
Other questions:
  • Suppose that the average speed (vrms) of carbon dioxide molecules (molar mass 44.0 g/mol) in a flame is found to be 2.67 105 m/s
    5·1 answer
  • Rearrange the momentum equation to solve for speed. Show all work.
    11·1 answer
  • What is the average acceleration during the time interval 0 seconds to 10 seconds?
    12·1 answer
  • Joe lives 100 miles away from Bill . What is Joes average speed if he reaches Bills home in50's?
    8·1 answer
  • An object starts from rest and accelerates at a rate of 4 m/s2 for 3 seconds. What is it's displacement from the start position?
    8·1 answer
  • NEED HELP ASAP, ILL GIVE YOU BRAINLIEST IF CORRECT (30POINTS)
    5·1 answer
  • the rock of 10 kg is falling near the Earth's surface.assume that g =10N/kg and the is no air resistance. what is the accelerati
    5·1 answer
  • Help please<br> It’s kinda urgent
    10·1 answer
  • Find the mass of a 165 N child
    8·1 answer
  • A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!