for a sateliite to have a speed of 7400 m/s how far must it be in metres above the earth's surface
V^2/(re+h)=9.8(re/(re+h))^2
solve for h.
M V^2/R = M*g*(Re/R)^2
(due to inverse square falloff in weight with R)
Re is the earth's radius = 6370*10^3 m.
R = Re^2*g/V^2 = 7.262*10^6 m = 7262 km
Distance above Earth = 7262 - 6370 km = 892 km
To determine how far a satellite must be above the Earth's surface to have a speed of 7400 m/s, we need to take into account the gravitational force exerted by the Earth. This force decreases as you move away from the Earth's surface.
To solve this, we can use the equation for the speed of a satellite in circular orbit:
v = √(GM/R)
where:
- v is the speed of the satellite,
- G is the gravitational constant (approximately 6.67430 × 10^(-11) N m^2/kg^2),
- M is the mass of the Earth (approximately 5.972 × 10^(24) kg),
- R is the distance of the satellite from the center of the Earth.
Rearranging the equation, we can solve for R:
R = GM/v^2
Plugging in the given values, we have:
R = (6.67430 × 10^(-11) N m^2/kg^2) * (5.972 × 10^(24) kg) / (7400 m/s)^2
Calculating this expression will give us the required distance, R.
R ≈ 4.232 × 10^7 meters
Therefore, the satellite must be approximately 4.232 × 10^7 meters (or 42,320 kilometers) above the Earth's surface to have a speed of 7400 m/s.