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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.

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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

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

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

Therefore, total energy of the spaceship

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

Energy required for launching the spaceship out of the solar system

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).

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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

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

$r=\sqrt{\frac{3Ene}{4\pi \rho g}}$

$=\sqrt{\frac{3×2.55×{10}^{4}×12×1.6×{10}^{-19}}{4×3.14×1.26×{10}^{3}×9.81}}$