All categories

1 year ago

combination or combinations of which method allow in principle measure the density of the extrasolar planet?

1 answer:

8 0

We can determine a planet's orbital period and separation from its star using any detection method. The transit method can yield sizes, whereas the astrometric and doppler approaches can provide minimum masses.

We can calculate average density by combining the transit and doppler approaches. Numerous physical properties, including the semi-major axis, stellar mass, star radius, planet radius, eccentricity, and inclination, are calculated from these observable data. The mass of the planet is also calculated using the star's combined radial velocity readings.

List briefly the planetary characteristics that, in theory, can be detected with the present detection techniques. We can determine a planet's orbital period and separation from its star using any detection method. The transit method can yield sizes, whereas the astrometric and doppler approaches can provide minimum masses.

To know more about orbital period

brainly.com/question/28068951

#SPJ4

You might be interested in

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4cm apart with a 28kV potential d

yulyashka [42]

Answer:

a) F = 3.2 10⁻¹⁰ N , b) v = 9.9 10⁷ m / s

Explanation:

a) The electric force is

F = q E

The electric field is related to the potential reference

V = E d

E = V / d

Let's replace

F = e V / d

Let's calculate

F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²

F = 3.2 10⁻¹⁰ N

b) For this part we can use kinematics

v² = v₀ + 2 a d

v = √ 2 ad

Acceleration can be found with Newton's second law

e V / d = m a

a = e / m V / d

a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²

a = 3,516 10⁻¹⁷ m / s²

Let's calculate the speed

v = √ (2 3,516 10¹⁷ 1.4 10⁻²)

v = √ (98,448 10¹⁴)

v = 9.9 10⁷ m / s

3 0

3 years ago

The bodies in this universe attract one another name the scientist who propounded this statement

Colt1911 [192]

<h2><em>Answer: According to wikipedia Sir Isaac Newton has told this statement.</em></h2><h2><em>hope its helps you.</em></h2><h2><em>have a great day.</em></h2><h2><em> keep smiling be happy stay safe. </em></h2><h2><em /></h2>

<em />

7 0

3 years ago

If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100

balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s = (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0

3 years ago

What are three (3) things you can do before exercise or sports that will keep you safe?

zaharov [31]

Answer: Drink water, practice, do some light stretches

Explanation:

3 0

3 years ago

If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?

HACTEHA [7]

$\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time, t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0 +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}$

5 0

2 years ago