Question

If an astronaut weighs 750 N on the ground, what will he weigh in earth orbit 400 km above the earth’s surface?

R_earth = 6.38 × 10^6 m,
m_earth=5.97×10^24 kg
G=6.674×10^−11 Nm^2/kg^2

Answer 1

The weight of the astronaut in Earth's orbit 400 km above the Earth's surface is 664.115 N

Step-by-step explanation:

The given parameters are;

The weight of the astronaut on the ground = 750 N

The height of the orbit above the Earths surface = 400 km

The radius of the Earth = 6.38 × 10⁶ m

The mass of the Earth = 5.97 × 10²⁴ kg

The universal gravitational constant G = 6.674 × 10⁻¹¹ Nm²/kg²

From Newton's law of universal gravitation, we have;

`Weight \ on \ ground \ of \ astronaut \ F_W =G* (M_(Earth) * m_(astronouat))/(R_(Earth)^(2)) = 750 \ N`

`F_W =6.674 * 10^(-11)* (5.97 * 10^(24)* m_(astronouat))/((6.38 * 10^6)^(2)) = 750 \ N`

`m_(astronaut)` = ((6.38 × 10⁶)² × 750)/((6.674 × 10⁻¹¹) × (5.97 × 10²⁴)) ≈ 76.62 kg

The mass of the astronaut ≈ 76.62 kg

For the weight of the astronaut in Earth orbit 400 km (400 × 10³ m) above the Earth's surface, we have;

`F_(W \ in \ orbit) =6.674 * 10^(-11)* (5.97 * 10^(24)* 76.62)/((6.38 * 10^6 + 400 * 10^3)^(2) ) = 664.115 \ N`

The weight of the astronaut in Earth's orbit 400 km above the Earth's surface = 664.115 N.