Question

The initial velocity $v_i$ required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity $v_e$ such that $v_i = \sqrt{\frac{x}{11}} v_e$. The value of x will be ________ .

Hint:

Remember that gravitational potential energy is $U = -GMm/r$, where r is the distance from the center of the Earth. A common mistake is to use the height 'h' instead of 'R+h'. Also, the escape velocity formula $v_e = \sqrt{2GM/R}$ is fundamental.

Answer 1

Correct Answer: 10

We will use the principle of conservation of mechanical energy.
The initial energy of the body on the Earth's surface is $E_i = KE_i + PE_i = \frac{1}{2}mv_i^2 - \frac{GMm}{R}$.
The final energy of the body at a height of 10R is $E_f = KE_f + PE_f = 0 - \frac{GMm}{R+10R} = -\frac{GMm}{11R}$. (At maximum height, final velocity is 0).
By conservation of energy, $E_i = E_f$:
$\frac{1}{2}mv_i^2 - \frac{GMm}{R} = -\frac{GMm}{11R}$.
$\frac{1}{2}mv_i^2 = \frac{GMm}{R} - \frac{GMm}{11R} = \frac{GMm}{R}\left(1 - \frac{1}{11}\right) = \frac{10}{11}\frac{GMm}{R}$.
$v_i^2 = \frac{20}{11}\frac{GM}{R}$.
The escape velocity from the Earth's surface is given by $v_e = \sqrt{\frac{2GM}{R}}$, so $v_e^2 = \frac{2GM}{R}$.
Substitute $v_e^2$ into the equation for $v_i^2$:
$v_i^2 = \frac{10}{11} \left(\frac{2GM}{R}\right) = \frac{10}{11} v_e^2$.
Taking the square root of both sides: $v_i = \sqrt{\frac{10}{11}} v_e$.
Comparing this with the given expression $v_i = \sqrt{\frac{x}{11}} v_e$, we find that x = 10.