Question

the radius of earth is about 6.38 x10^3 km. A 7.20 x10^3 N spacecraft travels away from earth. What is the weight of the spacecraft at the following distances from Earth's surface? a) 6.38 x 10^3 km

Answer 1

Answer:

$1796.65\ \text{N}$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

w = Weight of spracecraft at the surface = $7.2\times10^3\ \text{N}$

m = Mass of spracecraft

R = Radius of Earth = $6.38\times10^3\ \text{km}$

h = Elevation = $6.38\times10^3\ \text{km}$

G = Gravitational constant = $6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2$

M = Mass of Earth = $5.972\times 10^{24}\ \text{kg}$

$w=mg\\\Rightarrow m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{7.2\times 10^3}{9.81}\\\Rightarrow m=733.94\ \text{kg}$

From the gravitational law we have

$w'=\dfrac{GMm}{(r+h)^2}\\\Rightarrow w'=\dfrac{6.674\times10^{-11}\times 5.972\times 10^{24}\times 733.94}{(6.38\times10^6+6.38\times10^6)^2}\\\Rightarrow w'=1796.65\ \text{N}$

The weight of the spacecraft at the given height is $1796.65\ \text{N}$