Answer:
Speed of the satellite V = 6.991 × 10³ m/s
Explanation:
Given:
Force F = 3,000N
Mass of satellite m = 500 kg
Mass of earth M = 5.97 × 10²⁴
Gravitational force G = 6.67 × 10⁻¹¹
Find:
Speed of the satellite.
Computation:
Radius r = √[GMm / F]
Radius r = √[(6.67 × 10⁻¹¹ )(5.97 × 10²⁴)(500) / (3,000)
Radius r = 8.146 × 10⁶ m
Speed of the satellite V = √rF / m
Speed of the satellite V = √(8.146 × 10⁶)(3,000) / 500
Speed of the satellite V = 6.991 × 10³ m/s
i know can you plzz help me with this question im sorry i didt answer your question i just need hel.
Answer:
a) 0.167 μC/m^2
b) 1.887 * 10^4 V/m
Explanation:
Hello!
First let's find the surface charge density:
a)
Since thesatellite is metallic, the accumalted charge will be uniformly distribuited on its surface. Therefore the charge density σ will be:
σ = Q/A
Where A is the area of the satellite, which is:
A=4πr^2 = πd^2 = π(1.9m)^2
Therefore:
σ = (1.9)/(π (1.9)^2) μC/m^2 = 0.167 μC/m^2
Now let's calculate the electric field
b)
Just outside the surface of the satellite the elctric field will be:
E = σ/ε0
Where ε0=8.85×10^−12 C/Vm
Therefore:
E = (0.167*10^-6 C/m^2) / (8.85*10^-12 C/Vm) = 0.01887 *10^6 V/m
E = 1.887 * 10^4 V/m
No I don’t think so. But it worth a try tho. Try it out.
