A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from the Sun are r1 and r2, respectively. Find the relation between the time period of the planet in terms of r1 and r2.

Answers

Answer 1

Answer:

the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]

Explanation:

Given the data i  the question;

mass of sun = M

minimum and maximum distance = r1 and r2 respectively

Now, using Kepler's third law,

" the square of period T of any planet is proportional to the cube of average distance "

T² ∝ R³

average distance a = ( r1 + r2 ) / 2

we know that

T² = 4π²a³ / GM

T² = 4π² [( ( r1 + r2 ) / 2 )³ / GM ]

T² = 4π² [( ( r1 + r2 )³ / 8 ) / GM ]

T² = 4π² [( r1 + r2 )³ / 8GM ]

T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]

T = 2π √[( r1 + r2 )³ / 8GM ]

Therefore, the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]


Related Questions

Stars have different colors. What causes stars to have colors?

A. location

B. temperature

C. oxygen

D. carbon dioxide​

Answers

Answer:

temperature

Explanation:

temperature change forms different elements and different element sustain different colour

^temperature (sorry if its wrong!!!)

During a practice shot put throw, the 7.9-kg shot left world champion C. J. Hunter's hand at speed 16 m/s. While making the throw, his hand pushed the shot a distance of 1.4 m. Assume the acceleration was constant during the throw.

Required:
a. Determine the acceleration of the shot.
b. Determine the time it takes to accelerate the shot.
c, Determine the horizontal component of the force exerted on the shot by hand.

Answers

Answer:

a)   a = 91.4 m / s²,  b)    t = 0.175 s, c)  

Explanation:

a) This is a kinematics exercise

           v² = vox ² + 2a (x-xo)

           a = v² - 0/2 (x-0)

           

let's calculate

          a = 16² / 2 1.4

          a = 91.4 m / s²

b) the shooting time

          v = vox + a t

          t = v-vox / a

          t = 16 / 91.4

          t = 0.175 s

c) let's use Newton's second law

          F = ma

          F = 7.9 91.4

          F = 733 N

An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made from the power source and a 47.2 H inductor, determine the inductive reactance and the rms current through the inductor.

Answers

Answer:

The inductance is 17784.96 ohm and rms current is 4.77 mA.

Explanation:

Voltage, V = 120 V

frequency, f = 60 Hz

Inductance, L = 47.2 H

The rms  voltage is

[tex]V_{rms}=\frac{V_o}{\sqrt 2}\\\\V_{rms}=\frac{120}{\sqrt 2}\\\\V_{rms} = 84.87 V[/tex]

The reactance is given by

[tex]X_L = 2\pi f L\\\\X_L = 2\times 3.14\times 60\times 47.2 \\\\X_L = 17784.96 ohm[/tex]

The rms current is

[tex]I_{rms} =\frac{V_{rms}}{X_L}\\\\I_{rms}=\frac{84.87}{17784.96}\\\\I_{rms} = 4.77\times 10^{-3} A = 4.77 mA[/tex]

Put the following energy sublevels in order from least to greatest energy
A None of these
BIS. 25. 20, 35, 38, 34, 45, 46, 4d. 48
Cisas is4s, 20, 30, 40, 30, 40, 4f
D. is 25, 20, 35, 3p. 45, 3d, 4p, 40, 48

Answers

i would go with A here

What is true when an object floats in water? A. When an object floats, it exceeds the volume of water available. B. When an object floats, it displaces a volume of water equal to its own volume. C. When an object floats, it does not displace its entire volume.

Answers

Answer:

C. When an object floats, it does not displace its entire volume.

Explanation:

Buoyancy can be defined as an upward force which is created by the water displaced by an object.

According to Archimede's principle, it is directly proportional to the amount (weight) of water that is being displaced by an object.

Basically, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up. The buoyancy of an object is given by the formula;

[tex] Fb = pgV [/tex]

[tex] But, \; V = Ah [/tex]

[tex] Hence, \; Fb = pgAh [/tex]

Where;

Fb = buoyant force of a liquid acting on an object.

g = acceleration due to gravity.

p = density of the liquid.

v = volume of the liquid displaced.

h = height of liquid (water) displaced by an object.

A = surface area of the floating object.

The unit of measurement for buoyancy is Newton (N).

Additionally, the density of a fluid is directly proportional to the buoyant force acting on it i.e as the density of a liquid decreases, buoyancy decreases and vice-versa.

Furthermore, an object such as a boat, ship, ferry, canoe, etc, are able to float because the volume of water they displace weigh more than their own weight. Thus, if a boat or any physical object weighs more than the volume of water it displaces, it would sink; otherwise, it floats.

In conclusion, the true statement is that when an object floats, it does not displace its entire volume.

1. Compare and contrast the SI and the English systems of measurement.

Answers

Answer:The SI system is based on the number 10 as well as multiples and products of 10. This makes it much easier to use, and so it has been the accepted system in scientific and technical applications. The English system is more complicated as relationships between units of the same quantity aren't uniform.

Explanation:

Answer:

The metric system is an internationally agreed decimal system of measurement while The International System of Units (SI) is the official system of measurement in almost every country in the world

What quantity of heat is transferred when a 150.0g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of the heat flow?​

Answers

Answer:

Heat is flowing into the metal.

Explanation:

From the question given above, the following data were obtained:

Mass (M) of iron = 150 g

Initial temperature (T₁) = 25.0°C

Final temperature (T₂) = 73.3°C

Direction of heat flow =?

Next, we shall determine the change in the temperature of iron. This can be obtained as follow:

Initial temperature (T₁) = 25.0 °C

Final temperature (T₂) = 73.3 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 73.3 – 25

ΔT = 48.3 °C

Next, we shall determine the heat transfered. This can be obtained as follow:

Mass (M) of iron = 150 g

Change in temperature (ΔT) = 48.3 °C

Specific heat capacity (C) of iron = 0.450 J/gºC

Heat (Q) transfered =?

Q = MCΔT

Q = 150 × 0.450 × 48.3

Q = 3260.25 J

Since the heat transferred is positive, it means the iron metal is absorbing the heat. Thus, heat is flowing into the metal.

What is the biggest planet in the solar system

Answers

Answer:

Jupiter

Explanation:

Answer:

The answer is Jupiter.

Explanation:

Jupiter is an orange/yellow colored planet.

any one tell me about the earth rotation it rotatining or not with any proof? ​

Answers

The proof that the earth is rotating is the happens of night and day also the seasons, eg. winter, summer, autumn.

Oxygen is obtained through various methods. Which of the following methods involves a chemical
change?
1. Electrolysis of water
2. Distillation of liquid air
3. Heating of KCIO,
02
1 and 2
1 and 3

Answers

Answer:

1

Explanation:

Electrolysis is the passing of an current through a conducting solution, when the occurs, a chemical reaction takes place.

Heating a chemical will always cause a chemical reaction, which is why 3 is also correct

Some information as to why 2 is NOT correct.

2 is NOT a chemical reaction, but rather a process of physical separation. It uses selective boiling and condensation, but is not considered a chemical reaction.

as with 3, heating is not considered a chemical reaction, but rather a physical temperature change. This is always what it is considered to be (e.g boiling water is a physical temperature change, not a chemical reaction)

Hope this helps.

Hope this helps.

A heat engine exhausts 3 000 J of heat while performing 1 500 J of useful work. What is the efficiency of the engine

Answers

efficiency=work output/work input×100

since it exhausts(use up)3000j of heat that's the work input and the 1500j is the work input

efficiency=1500/3000×100

=50%

The diagram here shows an image being formed by a convex lens. Compared to the object at right, the image at left is-

larger and upright.

smaller and upright.

smaller and upside down.

larger and upside down.

Answers

Answer:

Smaller and upside down

Explanation:

To answer the question, we must recognise that the characteristics of the image formed by a convex lens depends on the position of the object from the lens.

From the diagram given in the question above, the following data were obtained:

1. The image is smaller than the object.

2. The image is inverted i.e upside down.

3. The image is closer to the lens

4. The image between 2f and f

Now, considering the options given in question above, the correct answer to the question is:

The image is smaller and upside down.

If an object travels on a circular path is an acceleration? What is changing to cause an acceleration?

Answers

For this reason, it can be safely concluded that an object moving in a circle at constant speed is indeed accelerating. It is accelerating because the direction of the velocity vector is changing.

To calculate the gravitational potential energy of a statue on a 10-meters-tall platform, you would have to know the statue's ______________

Answers

Answer:

mass

Explanation:

brainly A person's eye lens is 2.9 cm away from the retina. This lens has a near point of 25 cm and a far point at infinity. What must the focal length of this lens be in order for an object placed at the near point of the eye to focus on the retina

Answers

Answer: The focal length of the lens is 2.60 cm

Explanation:

The equation for lens formula follows:

[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]

where,

f = focal length = ? cm

v = image distance = 2.9 cm

u = Object distance = -25 cm

Putting values in above equation, we get:

[tex]\frac{1}{f}=\frac{1}{2.9}-\frac{1}{(-25)}\\\\\frac{1}{f}=\frac{1}{2.9}+\frac{1}{(25)}\\\\\frac{1}{f}=\frac{25+2.9}{2.9\times 25}\\\\f=\frac{72.5}{27.9}=2.60cm[/tex]

Hence, the focal length of the lens is 2.60 cm

Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car B is 100 from car A, car A begins to accelerate toward car B with a constant acceleration of 5 m/s/s. Let right be positive.
1) How much time elapses before the two cars meet? 2) How far does car A travel before the two cars meet? 3) What is the velocity of car B when the two cars meet?
4) What is the velocity of car A when the two cars meet?

Answers

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

[tex]A_a(t) = 5m/s^2[/tex]

To get the velocity, we integrate over time:

[tex]V_a(t) = (5m/s^2)*t + V_0[/tex]

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

[tex]V_a(t) = (5m/s^2)*t[/tex]

To get the position equation we integrate again over time:

[tex]P_a(t) = 0.5*(5m/s^2)*t^2 + P_0[/tex]

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

[tex]P_a(t) = 0.5*(5m/s^2)*t^2[/tex]

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

[tex]V_b(t) =20m/s[/tex]

To get the position equation, we can integrate:

[tex]P_b(t) = (20m/s)*t + P_0[/tex]

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

[tex]P_b(t) = (20m/s)*t + 100m[/tex]

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

[tex]P_a(t) = P_b(t)[/tex]

We can solve this for t.

[tex]0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0[/tex]

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

[tex]t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}[/tex]

We only care for the positive solution, which is:

[tex]t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s[/tex]

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

[tex]P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m[/tex]

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

[tex]V_a(t) = (5m/s^2)*11.48s = 57.4 m/s[/tex]

A ball on a frictionless plane is swung around in a circle at constant speed. The acceleration points in the same direction as the velocity vector.

a. True
b. False

Answers

I believe it is False, only because the plane is Frictionless. Hope this helps, good luck.

Answer:

False

Explanation:

You have a circle so think back to circular motion. Theres 2 directions, centripetal and tangential. The problem tells you there's a constant tangential speed so tangential acceleration is 0. However there is a centripetal acceleration acting on the ball that holds it in its circular motion (i.e. tension, or gravity). Since centripetal is perpendicular to the tangential direction, acceleration and velocity are in different directions.

Give an example of a substance with an amorphous structure.

Answers

Answer:

Tempered glass

Explanation:

When warmed, an amorphous substance has a non-crystalline architecture that differentiates from its isochemical liquid, but this does not go through structural breakdown or the glass transition.

Why is the temperature constant during the melting of water?​

Answers

[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}[/tex]

Why is the temperature constant during the melting of water?

[tex] \huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}[/tex]

[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{ REASON}}}}}}}[/tex]

THE HEAT WE R SUPPLYING TO THE WATER TO RAISE THE TEMP OF THE WATER IS USED BY THE MOLECULES TO BREAK INTERMOLECULAR BONDS WHICH HELP IN THE CHANGING OF THE LATTICE (STRUCTURE) OF THE WATER .

ICE HAS A HEXAGONAL RING LIKE STRUCTURE WHICH IS CONVERTED INTO REGULAR CRYSTALLINE STRUCTURE WHICH CAN ONLY BE FORMED WITH THE HELP OF FORMATION OF NEW BONDS AND BREAKDOWN OF OLDER ONES

THE AMOUNT OF ENERGY WHICH IS USED IN CONVERSATION OF THE STATE OF FROM SOLID TO LIQUID IS KNOWN AS LATENT HEAT OF FUSION.

SO TEMP REMAIN CONSTANT DURING CHANGE IN STATE .

[tex] \red \star{Thanks \: And \: Brainlist} \blue\star \\ \green\star If \: U \: Liked \: My \: Answer \purple \star[/tex]

g A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 3.00 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do

Answers

The desk is in equilbrium, so Newton's second law gives

F (horizontal) = p - f = 0

F (vertical) = n - mg = 0

==>   n = mg

==>   p = f = µn = µmg = 0.400 (80.0 kg) g = 313.6 N

The student pushes the desk 3.00 m, so she performs

W = (313.6 N) (3.00 m) = 940.8 Nm ≈ 941 J

of work.

A popular car stereo has four speakers, each rated at 60 W. In answering the following questions, assume that the speakers produce sound at their maximum power.
A) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.0 m.
B) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.5 m.
C) Find the intensity I of the sound waves produced by four 60-Wspeakers as heard by the driver. Assume that the driver is located 1.0 m from each of the two front speakers and 1.5 m from each of the two rear speakers.
D)The threshold of hearing is defined as the minimum discernible intensity of the sound. It is approximately 10^(-12) W/m2. Find the distance dfrom the car at which the sound from the stereo can still be discerned. Assume that the windows are rolled down and that each speaker actually produces 0.06 W of sound, as suggested in the last follow-up comment.

Answers

Answer:

Explanation:

Intensity of sound = sound energy emitted by source / 4 π d² , where d is distance of the source .

A )

Intensity of sound at 1 m distance = 60 /4 π d²

d = 1 m

Intensity of sound at 1 m distance = 60 /(4 π 1²)

= 4.78 W m⁻² s⁻¹

B )

Intensity of sound at 1.5 m distance = 60 /4 π d²

d = 1.5  m

Intensity of sound at 1 m distance = 60 /(4 π 1.5²)

= 2.12 W m⁻² s⁻¹

C )

Intensity of sound due to 4 speakers at 1.5 m distance = 4 x 60 /4 π d²

d = 1.5  m

= 4 x 60 /(4 π 1.5²)

= 8.48 W m⁻² s⁻¹

D )

Intensity of sound due to .06 W speaker must be 10⁻¹² W s ⁻² . Let the distance be d .

.06 /4 π d² = 10⁻¹²

d² = .06 /4 π 10⁻¹²

d = 6.9 x 10⁴ m .

Determine the absolute pressure on the bottom of a swimming pool 27.0 m by 8.9 m whose uniform depth is 1.8 m . Express your answer using two significant figures.

Answers

Answer:

[tex]P=17658Pa[/tex]

Explanation:

From the question we are told that:

Dimension

 [tex]L*B=27.0*8.9[/tex]

Depth [tex]d=1.8m[/tex]

Generally the equation for Volume of water is mathematically given by

 [tex]V=L*B*D[/tex]

 [tex]V=27.0*8.9*1.8[/tex]

 [tex]V=432.54m^3[/tex]

Therefore

Force at the bottom of the Pool

 [tex]F=\rho Vg[/tex]

Where

 [tex]\rho \ density\ of \ water(1000kg/m^3)[/tex]

 [tex]F=1000*432.54m^3*9.81[/tex]

 [tex]F=4.2*10^{6}N[/tex]

Generally the equation for Pressure at the bottom is mathematically given by

 [tex]P=\frac{Forece }{Area}[/tex]

 [tex]P=\frac{4.2*10^{6}N}{27.0*8.9}[/tex]

 [tex]P=17658Pa[/tex]

You work in the special effects department of a movie studio. You are currently working on a superhero movie where the hero is very
strong and cannot be hurt by normal weapons such as a sword. In the next scene, the villain
is going to hit the hero with a lead pipe.
The hero's costume is mostly made of foam and carbon so that is it very light and easy to move around in but will crumble if hit with
anything hard. For this scene, a section of the
costume needs to be replaced with a different material that will only dent, not crumble,
when hit with the pipe. You are on a deadline and look around the props
department for ideas, what could you use?
A. You find sheets of copper (Cu) used in an induced rescue factory scene
B. You find a barrel of phosphorus (P) used as a component in explosive powder
C. You find a tank of helium (He) used for balloons
D. You find large desk of Chlorine (Cl) used to kill bacteria and pools

Answers

Answer:

A. You find sheets of copper (Cu) used in an induced rescue factory scene

Explanation:

I would use a sheets of copper (Cu) used in an induced rescue factory scene because, copper is a metal and the only material out of the other options that would only dent and not crumble when hit by the lead pipe.

The other options in B - D are non-metals.

Phosphorus is a non-metal and is used as an explosive and thus cannot be used.

Helium is a gas used in filling balloons and thus cannot be used.

Chlorine is used in killing bacteria and thus cannot be used.

So, copper is the only option available since it is a metal and can thus accommodate a large force.

So, option A is the answer.

Could you show detailed steps in how to solve this problem please

Answers

Answer: See attached pic. Hope this helps.

Explanation:

The sound level measured in a room by a person watching a movie on a home theater system varies from 40 dB during a quiet part to 80 dB during a loud part. Approximately how many times louder is the latter sound

Answers

Answer:

[tex]\alpha=-3.01dB[/tex]

Explanation:

From the question we are told that:

Sound level intensity

 [tex]\triangle I=40dB-80dB[/tex]

Generally the equation for  intensity level  is mathematically given by

 [tex]\alpha=10log_{10}(I/I_x)dB[/tex]

Where

 I= Intensity measured

 [tex]I_x=Threshold\ of\ audibility[/tex]

 [tex]I_x= 10-12 W / m2[/tex]

 [tex]\alpha= 10 log10 \frac{I_1}{I_x} - 10 log10 \frac{}I_2{I_x}[/tex]

 [tex]\alpha= 10 log10 \frac{I_1}{I_2}[/tex]

 [tex]\alpha=10 log10\frac{40}{80}[/tex]

 [tex]\alpha=-3.01dB[/tex]

When the insulation resistance between a motor winding and the motor frame is tested, the value obtained is 1.0 megohm (106 Ω). How much current passes through the insulation of the motor if the test voltage is 1000 V?​

Answers

Answer:

0.001 A

Explanation:

Applying,

V = IR.............. Equation 1

Where V = Voltage of the motor, I = current, R = resistance

make I the subject of the equation

I = V/R.............. Equation 2

From the question,

Given: V = 1000 V, R = 1 MΩ = 10⁶ Ω

Substitute these values into equation 2

I = 1000/10⁶

I = 10⁻³ A

I = 0.001 A

What power (in kW) is supplied to the starter motor of a large truck that draws 260 A of current from a 25.5 V battery hookup

Answers

Answer:

P = 6.63 kW

Explanation:

Given that,

Current, I = 260 A

Voltage of the battery, V = 25.5 V

We need to find the power supplied to the starter motor. We know that,

P = VI

Put all the values,

P = 25.5 × 260

P = 6630 W

or

P = 6.63 kW

So, the power supplied to the motor is 6.63 kW.

Answer:

The power is 6.63 kW.

Explanation:

Current, I = 260  A

Voltage, V = 25.5 V

Power of an electrical appliance is given by

P = V I

P = 25.5 x 260

P = 6630 W

1 kW = 1000 W

So, the power is

P = 6.63 kW


If an electrical component with a resistance of 53 Q is connected to a 128-V source, how much current flows through the component?

Answers

Answer:

the current that flows through the component is 2.42 A

Explanation:

Given;

resistance of the electrical component, r = 53 Ω

the voltage of the source, V = 128 V

The current that flows through the component is calculated using Ohm's Law as demonstrated below;

[tex]V = IR\\\\I = \frac{V}{R} = \frac{128 \ V}{53 \ ohms} = 2.42 \ A[/tex]

Therefore, the current that flows through the component is 2.42 A


A current of 5.50 A flows in a conductor for 7.5 s. How much charge passes a given point in the conductor during this time?

Answers

56.1 neither songs were

A stopped organ pipe of length L has a fundamental frequency of 220 Hz. For which of the following organ pipes will there be a resonance if a tuning fork of frequency 660 Hz is sounded next to the pipe?

a. a stopped organ pipe of length L
b. a stopped organ pipe of length 2L
c. an open organ pipe of length L;
d. an open organ pipe of length 2L.

Answers

Answer:

Option (a), (d) are correct.

Explanation:

Frequency, f = 220 Hz

Resonant frequency = 660 Hz

The next frequency of stopped organ pipe is

2f, 3 f, 4 f ....

= 2 x 220 , 3 x 220 , 4 x 220 ....

= 440 Hz, 660 Hz, 880 Hz

So, the option (a) is correct.

The next resonant frequency of open organ pipe is

3 f, 5 f,...

= 3 x 220, 5 x 220 , ..

= 660 Hz, 1100 Hz,...

So, option (d) is correct.

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