Question

A rocket is launched with velocity 10km/s. If radius of earth is R, then maximum height attained by it will be?

Answer 1

Answer on Question #41798 – Physics – Physics, Mechanics | Kinematics | Dynamics

1. A rocket is launched with velocity $10\mathrm{km/s}$. If radius of earth is $R$, then maximum height attained by it will be?

$$v = 10^4 \frac{m}{s}$$ $$R = 6.37 \cdot 10^6 \, m$$ $$h - ?$$

**Solution.**

The maximum height attained by the rocket is determined by the law of conservation and transformation energy: the sum of the potential energy of the rocket (gravity energy) and its kinematic energy keeps constant.

If the Earth’s mass is $M$, then

$$- G \frac{mM}{R} + \frac{m v^2}{2} = - G \frac{mM}{R + h}.$$

The force of gravity at the surface of the Earth: $mg = G \frac{mM}{R^2}$.

One can find the height, at which the velocity equals to zero:

$$h = \frac{1}{\frac{1}{R} - \frac{v^2}{2GM}} - R, \quad h = \frac{1}{\frac{2g}{v^2} - \frac{1}{R}}.$$

Let check the dimension: $\left[h\right] = \frac{1}{m} = \frac{1}{\frac{s^2}{\left(\frac{m}{s}\right)^2} - \frac{1}{m}} = \frac{1}{m - \frac{1}{m}} = m.$

Let evaluate the quantity: $h = \frac{1}{\frac{2 \cdot 9.81}{(10^4)^2} - \frac{1}{6.37 \cdot 10^6}} = 2.55 \cdot 10^7 \, (m)$.

Answer: $2.55 \cdot 10^4 \, km$.

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