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
vlabodo [156]
3 years ago
13

Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W

hich best compares the satellites? Satellite X has a greater period and a faster tangential speed than Satellite Y. Satellite X has a greater period and a slower tangential speed than Satellite Y. Satellite X has a shorter period and a faster tangential speed than Satellite Y. Satellite X has a shorter period and a slower tangential speed than Satellite Y.
Physics
2 answers:
Mashcka [7]3 years ago
8 0

Answer:

B

Explanation:

jenyasd209 [6]3 years ago
4 0

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

T^{2}=\frac{4\pi^{2}}{GM}r^{3}    (1)

Where;

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24}kg is the mass of the Earth

r  is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}

V_{X}=18224.783 m/s (9)

<u>For Satellite Y:</u>

V_{Y}=\sqrt{\frac{GM}{r_{Y}}} (10)

V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}

V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

You might be interested in
Two astronauts are floating close to each other in space. Can they talk to each other without using any special device? plsss he
storchak [24]

Answer:

no they can't talk to each other bcoz of the lack of atmosphere.

Explanation:

l hope it helps you

5 0
3 years ago
Solve the inequality 2(n+3) – 4&lt;6. Then graph<br> the solution.
Aloiza [94]
The solution is 22 2(n+3)-4&6
6 0
3 years ago
Which term refers to energy derived from heat inside the earth
tino4ka555 [31]
I think the answer is Geothermal energy. 
3 0
3 years ago
A 1-kg rock is suspended by a massless string from one end of a
maxonik [38]

Answer:

The weight of measuring stick is 9.8 N

Explanation:

given information:

the mass of the rock, m_{r} = 1 kg

measuring stick, x =1 m

d = 0.25 m

to find the weight of measuring stick, we can use the following equation:

τ = Fd

τ = 0

F_{r} d - F_{s}d = 0

F_{r} = the force of the rock

F_{s} = the force of measuring stick

F_{s} =F_{r}

    = m g

    = 1 kg x 9.8 m/s

    = 9.8 N

thus, the weight of measuring stick is 9.8 N

6 0
3 years ago
An airplane flying parallel to the ground undergoes two consecutive displacements. The first is 50 km 60.0° west of north, and t
Rashid [163]

Answer:

LOL

Explanation:

IMAGINE POSTING UR CLASSWORK LOLL

8 0
2 years ago
Other questions:
  • Name two ways the eye adapts to dim light
    11·1 answer
  • Question points)
    11·1 answer
  • At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6
    13·1 answer
  • Which organism in the food chain is sometimes first level consumer and sometimes 2nd level
    15·1 answer
  • A quickly moving van and a slowly moving bird have a head-on collision. The bird splatters all over the windshield as a result.
    13·1 answer
  • Help please thank you whoever helps
    10·1 answer
  • Consider a helium balloon with volume 2.1 L.
    12·1 answer
  • 2 Pont
    10·1 answer
  • The work done to lift a 3kg mass to a<br> height of 10 m is
    7·1 answer
  • Which of the following correctly describes the law of conservation of matter?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!