A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4×1023 kg; radius of mars = 3395 km; G = 6.67×10-11 N m2 kg–2.

Asked by Pragya Singh | 1 year ago |  153

1 Answer

Solution :-

Speed of the rocket fired from the surface of mars ( v) = 2 km/s

Mass of the rocket = m

Mass of the Mars ( M) = 6.4 × 10²³ Kg

Radius of the Mars (R) = 3395 km = 3.395 × 106 m

Initial Potential energy =\( \dfrac{– GMm}{R}\)

Initial kinetic energy of the rocket = \( \dfrac{1}{2}\) mv²

Total initial energy = \( \dfrac{1}{2}\)mv² –\( \dfrac{GMm}{R}\)

Due to atmospheric resistance, 20% of the kinetic energy is lost by the rocket.

The remaining Kinetic energy= 80% of \( \dfrac{1}{2}\)mv²

\( \dfrac{2}{5}\) mv²= 0.4 mv² ——–(1)

When the rocket reaches the highest point, at the height h above the surface the kinetic energy will be zero and the potential energy is equal to 

\( \dfrac{– GMm}{R+h}\)

Applying Law of conservation of energy.

\( \dfrac{– GMm}{R+h}\) = \( \dfrac{– GMm}{R}\) + (0.4) mv²

\( \dfrac{– GMm}{(R+h)}\) = \( \dfrac{1}{R}\) [ GM – 0.4Rv2]

(R+\( \dfrac{h}{R}\)) = \( \dfrac{GM}{(GM – 0.4Rv^2)}\)

\( \dfrac{h}{R}\) = \( \dfrac{GM}{(GM – 0.4Rv^2)}\) – 1

\( \dfrac{h}{R}\)\( \dfrac{0.4Rv^2}{(GM – 0.4Rv^2)}\)

h =\( \dfrac{0.4R^2v^2}{(GM – 0.4Rv^2)}\)

h = \( \dfrac{0.4\times (2\times 10^{3})\times (3.395\times 10^{6})^{2}}{6.67\times 10^{-11}\times 6.4\times 10^{23}-0.4\times (2\times 10^{3})^{2}\times 3.395\times 10^{6}}\)

h = 495 km

Answered by Abhisek | 1 year ago

Related Questions

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 km; G = 6.67×10-11 N m2kg–2.

Class 11 Physics Gravitation View Answer

A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2×1030 kg).  

Class 11 Physics Gravitation View Answer

As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0×1024 kg, radius = 6400 km

Class 11 Physics Gravitation View Answer

Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

Class 11 Physics Gravitation View Answer

Two stars each of one solar mass (= 2×1030 kg) are approaching each other for a head-on collision. When they are a distance 10km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10km. Assume the stars to remain undistorted until they collide. (Use the known value of G).

Class 11 Physics Gravitation View Answer