Asatellite moves in a circular orbit around the earth at a speed of 6.9 km/s. determine the satellite's altitude above the surface of the earth. assume the earth is a homogeneous sphere of radius 6370 km and mass 5.98 × 1024 kg. the value of the universal gravitational constant is 6.67259 × 10−11 n · m2 /kg2 . answer in units of km.

Given: Velocity of satellite Vsat = 6.9 Km/s convert to m/s  Vsat = 6900 m/s

            Radius of Earth re = 6,370 Km;   Mass of Earth Me = 5.98 x 10²⁴ Kg

            Universal Gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Altitude of satellite r = ?

Formulas. F = ma;     a = V²/r       F = GMeMsat/r²

Equate for "r " from the three equations

GMeMsat/r² = MsatV²/r

r = GMe/V²

r = (6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)/6,900 m/s)²

r = 8,379,537.82 m or 8,379.53 Km

Deduct the the radius of our planet to find the altitude of the satellite

r = rsat - re 

r = 8, 379.59 Km - 6,370 Km

r = 2009.59 km or

r = 2010 Km altitude above the earth.

Ohm's law tells you that i = v/r where v is equivalent voltage and r is equivalent resistance of the circuit
since all circuits have the same voltage, your question is really asking which circuit has the lowest total resistance, since the lower the resistance, the higher the current
circuit a has a resistance of 5 ohms; circuit c is in series so has a resistance of 10 ohms; circuit b is in parallel and therefore has a resistance of 2.5 ohms (review your notes on adding resistances in parallel and series); therefore b has the highest current

Answer the radiant sun;