Electronics and Communication Engineering - Exam Questions Papers - Discussion
Discussion Forum : Exam Questions Papers - Exam Paper 10 (Q.No. 4)
4.
Find the electric charge required on the earth and moon to balance their gravitational attraction if the charge on the earth is 12.5 times that on the moon.
Mass of moon = 6.7 x 1022 Kg
Mass of earth = 6 x 1024 Kg
Distance between moon and earth = 380 Km
Universal gravitational constant = 6.7 x 10-11 Nm2/Kg2
Mass of moon = 6.7 x 1022 Kg
Mass of earth = 6 x 1024 Kg
Distance between moon and earth = 380 Km
Universal gravitational constant = 6.7 x 10-11 Nm2/Kg2
Answer: Option
Explanation:
Using law of gravitations, the force between two point masses is F = 
M1 = mass of moon = 6.7 x 1022 kg
M2 = mass of earth = 6 x 1024 kg
r = distance between masses = 380 km
G = universal gravitational constant
= 6.7 x 10- 11 Nm2/Kg2
Let Q1 and Q2 be the charge on moon and earth respectively. Now the gravitational force must be balanced by the force of repulsion.
Q1Q2 = 4pε0M1M2G
but Q2 = 12.5Q1
= 12.5 x 4pω0M1M2G
∴ Q2 = 6.11 x 1014
Q2 = 61.11 TC
Q1 = 4.89 TC.
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