Question

If it takes a planet 2.8 x 10 to orbit a star with a mass of 6.2x 10 kg, what is the average distance between the planet and the star

Answer 1

The answer can be found be using the formula of orbital velocity v=√{(G*M)/R}

Here G =Gravitational constant (G=6.67*10^-11 m³ kg⁻¹ s⁻²)

M= Mass of star

R = The distance between planet and the star

As we also know, v=ωR,

Here, ω= angular velocity

         R= radius

ωR=√{(G*M)/R}, now, ω=2π/T, where T is the orbital period of the planet

(2π/T)*R=√{(G*M)/R}, we "put to the power of 2" both sides to get rid of the square root and get:

{(2π/T)*R}²=[√{(G*M)/R}]², 

(4π²/T²)*R²=(G*M)/R, we multiply by R:

(4π²/T²)*R³=G*M, now we solve for R and get:

R³=(G*M*T²)/4π², we take the third root to get R:

R=∛{(G*M*T²)/4π²}, inserting the values of time and mass ( T= 2.8 x 10 s and Mass= 6.2x 10 kg)

R=∛{(G*M*T²)/4π²},

R=R=∛[{G=6.67*10^-11 x 6.2x 10 x (2.8 x 10^2)/4x3.14²},

R=9.36*10^11 m 

So this will be the average distance between the star and the planet. 

Answer 2

Answer:

9.36 * 10^11 m

Explanation:

The formula of general velocity can be used here =  v=√{(G*M)/R}.

G = gravitational constant or 6.67*10^-11 m³ kg⁻¹ s⁻²

M = Mass of the star

R = radius

To find v we use the formula v=ωR.

Here ω is the angular velocity; ω  = ω=2π/T

Here T is the orbital period of the planet

The formula is arranged likewise: (2π/T)*R=√{(G*M)/R}

Step 2: {(2π/T)*R}²=[√{(G*M)/R}]²

Step 3: (4π²/T²)*R²=(G*M)/R

Step 4: (4π²/T²)*R³=G*M

Step 5: R³=(G*M*T²)/4π²

Step 6: R=∛{(G*M*T²)/4π²}

Step 7: Add your numbers to the formula and we get the answer.