Answer:
7,790.38 m/s
Explanation:
Given the following :
What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 203 km above Earth's surface
Altitude = 203 km
Using the formula :
V = √GM/r
Where G = gravitational constant =6.67×10^-11
Kindly check attached picture for detailed explanation.
Answer:
The radius of orbit=
Explanation:
We are given that
Orbital speed=
m/s
We have to find the radius of orbit of spacecraft.
We know that
Gravitational constant=
Mass of earth=
Orbital speed=
Where G= Gravitational constant
M=Mass of earth
r=Radius of orbit
Substitute the values in the formula

Squaring on both sides



Hence, the radius of orbit=
Answer:
charge Qint = 7.17 10⁻⁴ C
Explanation:
For this problem we must use Gauss's law
F = ∫ E. dA = Qint / εₙ
let's form a Gaussian surface that is parallel to the surface, for example, a Cube. As the field is vertical and perpendicular to the surface, the field lines and the area vector are parallel whereby the scalar product is reduced to an ordinary product.
Φ = E A = Qint / ε₀
A = 1 km² (1000 m / 1km)² = 1 10⁶ m²
We can calculate the charge
Qint = E A ε₀
Qint = 81 1 10⁶ 8.85 10⁻¹²
Qint = 7.17 10⁻⁴ C