Question 282: Gravitation

Question

A body is projected vertically from earth's surface with $\left(\frac{1}{3}\right)^{\mathrm{rd}}$ of escape velocity. The maximum height reached by the body is ( $R=$ radius of earth)

Accepted Answer

Given:
The body is projected vertically upward with velocity $u = \dfrac{1}{3} v_{esc}$.
$R$ = radius of Earth.
Step 1: Escape velocity from Earth
Escape velocity from Earth is given by:
$$ v_{esc} = \sqrt{\dfrac{2GM}{R}} $$
Step 2: Initial Kinetic Energy and Potential Energy
At EarthÃ¢ÂÂs surface:
Potential Energy: $U_1 = -\dfrac{GMm}{R}$
Kinetic Energy: $K_1 = \dfrac{1}{2} m u^2$
Total initial energy:
$$ E_1 = K_1 + U_1 = \dfrac{1}{2}m u^2 - \dfrac{GMm}{R} $$
Step 3: At Maximum Height $h$
At maximum height $h$:
Final velocity = $0$ (momentarily stops)
Distance from centre = $R + h$
Total energy at top:
$$ E_2 = 0 - \dfrac{GMm}{R+h} = -\dfrac{GMm}{R+h} $$
Step 4: Conservation of Mechanical Energy
By conservation of energy,
$$ E_1 = E_2 $$
Substitute values:
$$ \dfrac{1}{2} m u^2 - \dfrac{GMm}{R} = -\dfrac{GMm}{R+h} $$
Step 5: Substitute $u = \dfrac{1}{3} v_{esc}$
Recall $v_{esc}^2 = \dfrac{2GM}{R}$
So:
$$ u^2 = \left(\dfrac{1}{3} v_{esc}\right)^2 = \dfrac{1}{9} v_{esc}^2 = \dfrac{1}{9}\left(\dfrac{2GM}{R}\right) = \dfrac{2GM}{9R} $$
So,
$$ \dfrac{1}{2} m u^2 = \dfrac{1}{2} m \left(\dfrac{2GM}{9R}\right) = \dfrac{GMm}{9R} $$
Plug this into the energy equation:
$$ \dfrac{GMm}{9R} - \dfrac{GMm}{R} = -\dfrac{GMm}{R+h} $$
$-\dfrac{GMm}{R} + \dfrac{GMm}{9R} = -\dfrac{GMm}{R+h}$
Combine:
$$ \dfrac{GMm}{9R} - \dfrac{GMm}{R} = -\dfrac{GMm}{R+h} $$
$$ \dfrac{GMm}{9R} - \dfrac{9GMm}{9R} = -\dfrac{GMm}{R+h} $$
$$ \dfrac{GMm - 9GMm}{9R} = -\dfrac{GMm}{R+h} $$
$$ \dfrac{-8GMm}{9R} = -\dfrac{GMm}{R+h} $$
Step 6: Simplify and Solve for $h$
Remove minus signs and $GMm$:
$$ \dfrac{8}{9R} = \dfrac{1}{R+h} $$
Take reciprocal:
$$ R + h = \dfrac{9R}{8} $$
$$ h = \dfrac{9R}{8} - R = \dfrac{9R - 8R}{8} = \dfrac{R}{8} $$
Final Answer:
$ \boxed{\frac{R}{8}} $
So, the correct option is
Option B
.

Answer

$\frac{\mathrm{R}}{4}$

Answer

$\frac{\mathrm{R}}{8}$

Answer

$\frac{R}{9}$

Answer

$\frac{R}{6}$

Gravitation

A body is projected vertically from earth's surface with \left(\frac{1}{3}\right)^{\mathrm{rd}} of escape velocity. The maximum height reached by the body is ( R= radius of earth)

Choose the correct answer: