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?

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

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; link

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

radiation

explanation:

heat is transmitted by radiation