Question

A rocket is fired vertically from the ground with a resultant vertical acceleration of 10 m s$^{-2}$. The fuel is finished in 1 minute and it continues to move up. The approximate maximum height reached by rocket from the surface of Earth is (mass of Earth = 6 × 10$^{24}$ kg; G = 6.6 × 10$^{-11}$ N m$^2$ kg$^{-2}$; radius of Earth = 6.4 × 10$^6$ m)

• A. 39 km
• B. 48 km
• C. 52 km
• D. 59 km

Answer 1

Answer: (A)

39 km

Answer 2

The problem asks for the approximate maximum height reached by a rocket fired vertically from the ground. The motion can be divided into two phases:

Phase 1: Accelerated motion (fuel burning)

The rocket starts from rest ($u_1 = 0$) with a resultant vertical acceleration ($a = 10 \, \text{m s}^{-2}$) for a duration of $t = 1 \, \text{minute} = 60 \, \text{s}$.

1. Height covered in Phase 1 ($h_1$): Using the kinematic equation $h_1 = u_1 t + \frac{1}{2}at^2$: $h_1 = (0)(60) + \frac{1}{2}(10 \, \text{m s}^{-2})(60 \, \text{s})^2$ $h_1 = \frac{1}{2}(10)(3600)$ $h_1 = 5 \times 3600 = 18000 \, \text{m} = 18 \, \text{km}$

2. Velocity at the end of Phase 1 ($v_1$): Using the kinematic equation $v_1 = u_1 + at$: $v_1 = 0 + (10 \, \text{m s}^{-2})(60 \, \text{s})$ $v_1 = 600 \, \text{m s}^{-1}$

Phase 2: Motion under gravity (fuel exhausted)

After 1 minute, the fuel is finished, and the rocket continues to move upwards under the influence of Earth's gravity. The initial velocity for this phase is $v_1 = 600 \, \text{m s}^{-1}$ from a height $h_1 = 18 \, \text{km}$ above the ground. The rocket will reach its maximum height when its final velocity becomes zero.

Since the height reached is relatively small compared to the Earth's radius ($R_E = 6.4 \times 10^6 \, \text{m} = 6400 \, \text{km}$), we can approximate the acceleration due to gravity ($g$) as constant ($g \approx 9.8 \, \text{m s}^{-2}$). The problem asks for an "approximate" height, which supports this assumption.

1. Additional height covered in Phase 2 ($h_2$): Using the kinematic equation $v_f^2 = v_i^2 - 2gh_2$, where $v_f = 0$ (at maximum height) and $v_i = v_1$: $0^2 = v_1^2 - 2gh_2$ $h_2 = \frac{v_1^2}{2g}$ Using $g = 9.8 \, \text{m s}^{-2}$: $h_2 = \frac{(600 \, \text{m s}^{-1})^2}{2 \times 9.8 \, \text{m s}^{-2}} = \frac{360000}{19.6}$ $h_2 \approx 18367.35 \, \text{m} \approx 18.37 \, \text{km}$

2. Total maximum height ($H_{max}$): The maximum height reached from the surface of Earth is the sum of the heights from both phases: $H_{max} = h_1 + h_2$ $H_{max} = 18 \, \text{km} + 18.37 \, \text{km}$ $H_{max} = 36.37 \, \text{km}$

This value is approximately $36 \, \text{km}$. Looking at the options, $39 \, \text{km}$ is the closest. The difference might be due to rounding $g$ to $10 \, \text{m s}^{-2}$ or other approximations.

Let's check with $g = 10 \, \text{m s}^{-2}$ for the second phase, as sometimes done in problems with round numbers: $h_2 = \frac{(600 \, \text{m s}^{-1})^2}{2 \times 10 \, \text{m s}^{-2}} = \frac{360000}{20} = 18000 \, \text{m} = 18 \, \text{km}$ In this case, $H_{max} = 18 \, \text{km} + 18 \, \text{km} = 36 \, \text{km}$.

The given options are quite spaced out. The closest option is 39 km. The small difference could be attributed to using a slightly different value for $g$ (e.g., if the problem implicitly assumes $g$ is slightly less than 9.8 or 10, or if the "approximate" nature allows for some deviation). However, using the provided values for G, M_E, R_E, we can calculate $g_0 = \frac{GM_E}{R_E^2} = \frac{6.6 \times 10^{-11} \times 6 \times 10^{24}}{(6.4 \times 10^6)^2} = \frac{39.6 \times 10^{13}}{40.96 \times 10^{12}} = \frac{396}{40.96} \approx 9.667 \, \text{m s}^{-2}$. If we use $g = 9.667 \, \text{m s}^{-2}$: $h_2 = \frac{360000}{2 \times 9.667} = \frac{360000}{19.334} \approx 18629 \, \text{m} \approx 18.63 \, \text{km}$ $H_{max} = 18 \, \text{km} + 18.63 \, \text{km} = 36.63 \, \text{km}$.

All constant $g$ approximations lead to a value around 36-37 km. The closest option is 39 km. It's possible the exact value of $g$ or the "approximate" nature of the question allows for this difference.

Let's consider the energy conservation approach to see if it yields a significantly different answer, although for these heights, constant g is usually a good approximation. Initial total energy (at $R_E + h_1$ with velocity $v_1$): $E_i = \frac{1}{2}mv_1^2 - \frac{GM_E m}{R_E+h_1}$ Final total energy (at $R_E + H_{max}$ with velocity 0): $E_f = -\frac{GM_E m}{R_E+H_{max}}$ Equating $E_i = E_f$: $\frac{1}{2}v_1^2 - \frac{GM_E}{R_E+h_1} = -\frac{GM_E}{R_E+H_{max}}$ $\frac{1}{R_E+H_{max}} = \frac{1}{R_E+h_1} - \frac{v_1^2}{2GM_E}$ $R_E+h_1 = 6.4 \times 10^6 + 18000 = 6.418 \times 10^6 \, \text{m}$ $GM_E = 6.6 \times 10^{-11} \times 6 \times 10^{24} = 39.6 \times 10^{13} \, \text{N m}^2 \text{kg}^{-1}$ $\frac{v_1^2}{2GM_E} = \frac{(600)^2}{2 \times 39.6 \times 10^{13}} = \frac{3.6 \times 10^5}{79.2 \times 10^{13}} = \frac{3.6}{79.2} \times 10^{-8} = 0.04545 \times 10^{-8} = 4.545 \times 10^{-10}$ $\frac{1}{R_E+h_1} = \frac{1}{6.418 \times 10^6} \approx 0.1558 \times 10^{-6}$ $\frac{1}{R_E+H_{max}} = 0.1558 \times 10^{-6} - 0.0004545 \times 10^{-6} = (0.1558 - 0.0004545) \times 10^{-6} = 0.1553455 \times 10^{-6}$ $R_E+H_{max} = \frac{1}{0.1553455 \times 10^{-6}} \approx 6.437 \times 10^6 \, \text{m}$ $H_{max} = (6.437 \times 10^6 \, \text{m}) - (6.4 \times 10^6 \, \text{m}) = 0.037 \times 10^6 \, \text{m} = 37 \, \text{km}$.

The energy conservation method gives approximately $37 \, \text{km}$. This is still closest to $39 \, \text{km}$. The small difference might be due to the "approximate" nature of the question or slight rounding in the options.

The most precise calculation (using energy conservation and given constants) gives $37 \, \text{km}$. Given the options, $39 \, \text{km}$ is the most reasonable approximate answer.