Astronomers are comparing two stars that are known to have the same luminosity. Star A is observed to be 4 times brighter than s
1 answer:
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Answer:
W = 28226.88 N
Explanation:
Given,
Mass of the satellite, m = 5832 Kg
Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m
The time period of the orbit, T = 1.9 h
= 6840 s
The radius of the planet, R = 4.38 x 10⁶ m
The time period of the satellite is given by the formula
second
Squaring the terms and solving it for 'g'
g = 4 π² m/s²
Substituting the values in the above equation
g = 4 π²
g = 4.84 m/s²
Therefore, the weight
w = m x g newton
= 5832 Kg x 4.84 m/s²
= 28226.88 N
Hence, the weight of the satellite at the surface, W = 28226.88 N