Answer:
A) 2.650 km
Explanation:
The relationship between acceleration of gravity and gravitational constant is:
[tex]g = \frac{Gm}{R^2}[/tex] ---- (1)
Where
[tex]R = 6,400 km[/tex] -- Radius of the earth.
From the question, we understand that the gravitational field of the rocket is 50% of its original value.
This means that:
[tex]g_{rocket} = 50\% * g[/tex]
[tex]g_{rocket} = 0.50 * g[/tex]
[tex]g_{rocket} = 0.5g[/tex]
For the rocket, we have:
[tex]g_{rocket} = \frac{Gm}{r^2}[/tex]
Where r represent the distance between the rocket and the center of the earth.
Substitute 0.5g for g rocket
[tex]0.5g = \frac{Gm}{r^2}[/tex] --- (2)
Divide (1) by (2)
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}/\frac{Gm}{r^2}[/tex]
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}*\frac{r^2}{Gm}[/tex]
[tex]\frac{1}{0.5} = \frac{1}{R^2}*\frac{r^2}{1}[/tex]
[tex]2 = \frac{r^2}{R^2}[/tex]
Take square root of both sides
[tex]\sqrt 2 = \frac{r}{R}[/tex]
Make r the subject
[tex]r = R * \sqrt 2[/tex]
Substitute [tex]R = 6,400 km[/tex]
[tex]r = 6400km * \sqrt 2[/tex]
[tex]r = 6400km * 1.414[/tex]
[tex]r = 9 049.6\ km[/tex]
The distance (d) from the earth surface is calculated as thus;
[tex]d = r - R[/tex]
[tex]d = 9049.6\ km - 6400\ km[/tex]
[tex]d = 2649.6\ km[/tex]
[tex]d = 2650\ km[/tex] --- approximated
