Turn off lights when leaving rooms.
Unplug unused appliances. Even when powered off these appliances use electricity.
Replace regular light bulbs with energy saving bulbs.
Gravitational force is given by, 
Where, m and M are the masses of the objects, R is the distance between them and G gravitational constant.
Gravitational force of the star on planet 1, 
Gravitational force of the star on planet 2, 
Ratio, 

Therefore, the gravitational force of the star on the planet 1 is three times that on planet 2.
Answer:
995.12 N/C
Explanation:
R = 9 cm = 0.09 m
σ = 9 nC/m^2 = 9 x 10^-9 C/m^2
r = 9.1 cm = 0.091 m
q = σ x 4π R² = 9 x 10^-9 x 4 x 3.14 x 0.09 x 0.09 = 9.156 x 10^-10 C
E = kq / r^2
E = ( 9 x 10^9 x 9.156 x 10^-10) / (0.091 x 0.091)
E = 995.12 N/C
Answer:
r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s
Answer:
The acceleration would be 3.455.