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Q8.25:
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 = $6.4 \times 10^{23}$ kg; radius of mars = 3395 km; G = $6.67 \times 10^{-11}$ N m$^2$ kg$^{–2}$.
Solution :
Step 1: Given Variables and Constants
- Initial velocity of the rocket, $v = 2 \times 10^3 \, \text{m s}^{-1}$
- Mass of Mars, $M = 6.4 \times 10^{23} \, \text{kg}$
- Radius of Mars, $R = 3395 \times 10^3 \, \text{m}$
- Gravitational Constant, $G = 6.67 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2}$
- Energy loss factor, $\eta = 20\% = 0.2$
Step 2: Energy Conservation Analysis
The total initial mechanical energy ($E_i$) of the rocket at the surface consists of kinetic energy ($K$) and potential energy ($U$). [Per the Principle of Conservation of Energy].
$E_i = K + U = \frac{1}{2}mv^2 - \frac{GMm}{R}$
Since 20% of the initial kinetic energy is lost to atmospheric resistance, the effective energy available at the surface to reach the maximum height $h$ is:
$E_{effective} = (K - 0.2K) + U = 0.8 \left( \frac{1}{2}mv^2 \right) - \frac{GMm}{R}$
$E_{effective} = 0.4mv^2 - \frac{GMm}{R}$
Step 3: Energy at Maximum Height
At the maximum height ($R+h$), the velocity of the rocket is zero. Therefore, the total energy ($E_f$) at that point is purely potential:
$E_f = -\frac{GMm}{R+h}$
Equating $E_{effective}$ to $E_f$ [By the Work-Energy Theorem]:
$0.4mv^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}$
Dividing by mass $m$ (as it cancels out):
$0.4v^2 - \frac{GM}{R} = -\frac{GM}{R+h}$
Step 4: Numerical Computation
Calculating the individual components:
- $0.4v^2 = 0.4 \times (2000)^2 = 0.4 \times 4 \times 10^6 = 1.6 \times 10^6 \, \text{J kg}^{-1}$
- $\frac{GM}{R} = \frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{3395 \times 10^3} \approx 12.58 \times 10^6 \, \text{J kg}^{-1}$
Rearranging for $R+h$:
$\frac{GM}{R+h} = \frac{GM}{R} - 0.4v^2$
$\frac{GM}{R+h} = 12.58 \times 10^6 - 1.6 \times 10^6 = 10.98 \times 10^6$
$R+h = \frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{10.98 \times 10^6} \approx 3.887 \times 10^6 \, \text{m}$
Step 5: Final Displacement
$h = (R+h) - R = 3887 \, \text{km} - 3395 \, \text{km} = 492 \, \text{km}$
Final Solution: The rocket will reach a maximum distance of 492 km from the surface of Mars.
More Questions from Class 11 Physics Gravitation Additional Exercises
- Q8.22: 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 = $6.0 \times 10^{24}$ kg, radius = 6400 km.
- Q8.23: 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 = $2 \times 10^{30}$ kg).
- Q8.24: 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 = $2 \times 10^{30}$ kg; mass of mars = $6.4 \times 10^{23}$ kg; radius of mars = 3395 km; radius of the orbit of mars = $2.28 \times 10^8$ km; G = $6.67 \times 10^{-11}$ N m$^2$ kg$^{–2}$.
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