Q: A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun = 2 × 1030 kg; mass of mars = 6.4 × 1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 108 kg; G= 6.67 × 10–11 N m2 kg–2.

More questions like this

Answer:

Given,

Mass of the spaceship, m = 1000 kg

Mass of the Sun, MS = 2 × 1030 kg

Mass of Mars, MM = 6.4 × 1023 kg

Orbital radius of Mars, R = 2.28 × 108 km =2.28 × 1011 m

Radius of Mars, r = 3395 km = 3.395 × 106 m

Universal gravitational constant, G= 6.67 × 10–11 N m2 kg–2

Potential energy of the spaceship due to the gravitational attraction of the Sun

VS=- GMS mR

Potential energy of the spaceship due to the gravitational attraction of Mars

VM=- GMM mr

Since, the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be zero.

Therefore, total energy of the spaceship

V=VS+VM==- GMS mR- GMM mr=- GmMSR+MMr

The negative sign indicates that the system is in bound state.

Energy required for launching the spaceship out of the solar system

=  (Total energy of the spaceship)

= GmMSR+MMr

=6.67 ×1011×1000×2 × 10302.28 × 1011 +6.4 × 10233.395 × 106

=6.67 ×108×87.72 × 1017+1.88 × 1017

=6.67 ×108×89.50 × 1017

=598× 109 J

 6× 1011 J

Q: An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C−1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm−3. Estimate the radius of the drop. (g = 9.81 m s−2; e = 1.60 × 10−19 C).

More questions like this

Answer:

Since the drop is stationary, force (FE) due to electric field E is equal to the weight of the oil drop (W)

Let the radius of the oil drop be r

Mass of the oil drop m = Volume of the drop × Density of oil

m =43πr3×ρ 

W=mg=43πr3×ρ×g  

Electrostatic Force FE = qE

Excess electrons on an oil drop, n = 12

Charge on an electron, e = 1.6 × 10−19 C

q = Net charge on the oil drop = n e

= 12 × 1.6 × 10−19 C

Electric field intensity, E = 2.55 × 104 N C−1

Density of oil, ρ = 1.26 gm/cm3 = 1.26×103 kg/m3

Acceleration due to gravity, g = 9.81 m s−2

Putting the values in

E n e=43πr3×ρ×g 

r=3Ene4πρg3

=3×2.55×104×12×1.6×10-194×3.14×1.26×103×9.813

=946.09×10-213=9.82×10-7 m