Question

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A in a circular orbit of radius 5000 km and period 4.0 hours. What is the radius of Planet R47A

Answer 1

The radius of Planet R47A is approximately 11,459 km.

Step-by-step explanation:

To find the radius of Planet R47A, we can use Kepler's Third Law which relates the period and radius of an orbit. Kepler's Third Law states that the square of the period is proportional to the cube of the radius. So we have:

T^2 ∝ R^3

Given that the period is 4.0 hours and the radius of the orbit is 5000 km, we can now solve for the radius of Planet R47A. Let's rearrange the equation:

R = (T^2 * GM / 4π^2)^(1/3)

Plugging in the values, we get:

R = (4.0^2 * G * M / (4π^2))^(1/3)

Since the gravitational acceleration is given as 3.45 m/s^2, we can use the formula for gravitational acceleration:

GM = g * R^2

Plugging in the values, we get:

(3.45 * 5000^2)^(1/3)

Simplifying, we find that the radius of Planet R47A is approximately 11,459 km.

Answer 2

`2.6×10^6\:\text{m}`

Explanation:

The acceleration due to gravity g is defined as

`g = G(M)/(R^2)`

and solving for R, we find that

`R = \sqrt{(GM)/(g)}\:\:\:\:\:\:\:(1)`

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration
`F_c` experienced by the satellite is equal to the gravitational force
`F_G` or

`F_c = F_G \Rightarrow m(v^2)/(r) = G(mM)/(r^2)\:\:\:\:\:(2)`

The orbital velocity v is the velocity of the satellite around the planet defined as

`v = (2\pi r)/(T)`

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

`(4\pi^2 r)/(T^2) = G(M)/(r^2)`

Solving for M, we get

`M = (4\pi^2 r^3)/(GT^2)`

Putting this expression back into Eqn(1), we get

`R = \sqrt{(G)/(g)\left((4\pi^2 r^3)/(GT^2)\right)}`

`\:\:\:\:=(2\pi)/(T)\sqrt{(r^3)/(g)}`

`\:\:\:\:=\frac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\frac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}`

`\:\:\:\:= 2.6×10^6\:\text{m}`