Question 6: Gravitation

Question

What is the maximum height attained by a body projected with a velocity equal to one-third of the escape velocity from the surface of the earth? (Radius of the earth $$=R$$ )

Accepted Answer

From conservation of energy,
$$\frac{1}{2} m v_1^2-\frac{G M_e m}{R}=\frac{1}{2} m v_2^2-\frac{G M_e m}{R+H}$$
Here, $$v_2=0$$ (At maximum height)
$$	herefore \quad \frac{1}{2} m v_1-\frac{G M_e m}{R}=-\frac{G M_e m}{R+H}$$
Solving, we get
$$H=\frac{R}{\frac{2 g R}{v_1^2}-1}=\frac{R}{\frac{v_e^2}{v_1^2}-1}$$
Where $$v_e$$ is escape velocity
Here $$v=\frac{1}{3} v_e \Rightarrow \frac{v_e}{v}=3$$
$$	herefore \quad H=\frac{R}{9-1}=\frac{R}{8}$$

Answer

$$R / 2$$

Answer

$$R / 3$$

Answer

$$R / 5$$

Answer

$$R / 8$$

Gravitation

What is the maximum height attained by a body projected with a velocity equal to one-third of the escape velocity from the surface of the earth? (Radius of the earth $=R$ )

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