Question

An intergalactic spaceship arrives at a distant planet that rotates on its axis with a period of ?f=T%3D26%20hours" id="TexFormula1" title="T=26 hours" alt="T=26 hours" align="absmiddle" class="latex-formula">. The mass of the planet is $M=4.9*10^{25} kg$. The spaceship enters a circular orbit with an orbital period that is equal to the planet's period for the rotation about its axis, T.
A) Enter an expression for the radius of the spaceship's orbit.
(I tried $R=(T^{2} GM/4pi^{2} )^{1/3}$ but its says incorrect and that's all I've been able to find)
B) Calculate the orbital radius in meters.

Answer 1

The orbital period could be obtained from the question as 8.98 * 10^7 m.

What is the orbital period?

We know that the solar system is composed of the sun and the planets. The planets are known to move around the sun in concentric circles. Following the heliocentric model of the solar system, the sun is at the center of the solar system at all times and all the other planets tend to move round the sun at a good distance that is appropriate for each.

Now we have that;

T = √4π^2r^3/GM

T = period of the orbit

r = radius of the orbit

G = gravitational constant

M = mass of the planet

Hence, we have;

R = ∛T^2GM/4π^2

R =∛ (26 * 60 * 60)^2 * 6.67 * 10^-11 * 4.9 * 10^25/ 4 * (3.142)^2

R = ∛2.86 * 10^25/39.49

R = 8.98 * 10^7 m

Learn more about orbital period:brainly.com/question/14494804

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