Question

An artificial satellite is in a circular orbit around a planet of radius r = 5.25 Ã 10^3 km at a distance d = 450.0 km from the planet's surface. The period of revolution of the satellite around the planet is T = 2.15 hours. What is the average density of the planet?

Answer 1

Answer:

3020.68 kg/m^3

Explanation:

r = 5.25 x 10^3 km = 5.25 x 10^6 m, d = 450 km = 450 x 10^3 m

T = 2.15 hours = 2.15 x 3600 second = 7740 second

Let the density of the planet is ρ and M be the mass of planet.

The formula for the orbital velocity is

$v = \sqrt{\frac{GM}{r+d}}$

Time period is given by

$T = {\frac{2\pi (r+d)}{v}}$

$T = \frac{2\pi (r +d)^{1.5}}{\sqrt{GM}}$

$7740= \frac{2\pi (5700\times 1000)^{1.5}}{\sqrt{6.67\times 10^{-11}M}}$

M = 1.83 x 10^24 kg

Density = mass / Volume

ρ = 1.83 x 10^24 / (4/3 x 3.14 x (5.25 x 10^6)^3)

ρ = 3020.68 kg/m^3

Answer 2

The average density of the planet is; 3020.68 kg/m³

What is average density?

We are given;

Radius; r = 5.25 × 10³ km = 5.25 × 10⁶ m

Distance; d = 450 km = 450 × 10³ m

Period; T = 2.15 hours = 7740 secs

The formula for the orbital velocity is;

V = √(GM/(r + d))

Formula for time period is;

T = 2π(r + d)/v

Thus;

T = 2π(r + d)/√(GM/(r + d))

T = 2π((r + d)^1.5)/√GM

Where G = 6.67 * 10⁻¹¹ N.m²/kg²

Plugging other relevant values gives us;

M = 1.83 * 10²⁴ kg

Formula for density is;

Density = mass/volume

Earth is spherical and as such, Volume = 4/3 πr³.

Thus;

Density = (1.83 * 10²⁴)/((4/3) * π * (5.25 × 10⁶)³

Density = 3020.68 kg/m³

Read more about Average Density at; https://brainly.com/question/1354972