Question

A satellite 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.

Answer 1

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.

Answer 2

The altitude of the satellite above the surface of the Earth is 1,528.48 km.

The given parameters;

• speed of the satellite, v = 6.9 km/s
• radius of the Earth, R = 6370 km
• mass of the Earth of the, m = 5.98 x 10²⁴ kg
• gravitational constant, G = 6.67259 x 10⁻¹¹ Nm²/kg²

The altitude of the satellite above the surface of the Earth is calculated as follows;

$F = \frac{mv^2}{r} = \frac{GmM}{R^2} \\\\\frac{v^2}{r} = \frac{GM}{R^2} \\\\r = \frac{v^2 R^2}{GM} \\\\r = \frac{(6900)^2 \times (6,370,000)^2}{(6.67259\times 10^{-11}) \times (5.98\times 10^{24})} \\\\r = 4,841,516.6 \ m\\\\r = 4,841.52 \ km$

The altitude of the satellite = 6370 km - 4,841.52 km = 1,528.48 km

Thus, the altitude of the satellite above the surface of the Earth is 1,528.48 km.

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