Question 5: Gravitation

Question

A space ship is launched into a circular orbit close to earth’s surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull?
(Radius of earth = 6400 $km$, g = $9.8 m/s^{2}$)

Accepted Answer

The orbital velocity of space ship in
circular
$v_0=\sqrt{\frac{G M}{r}} \$
If space ship is very close to earth surface,
r = radius of earth = R
$\ \begin{aligned}
& 	herefore \quad v_0=\sqrt{\frac{G M}{R}}=\sqrt{\frac{R^2 g}{R}}=\sqrt{R g} \
&\left(	ext { since } g=\frac{G M}{R^2}ight)
\end{aligned}$
$ \ \begin{aligned}
	ext { Here, } \quad R & =6400 \mathrm{~km}=6.4 	imes 10^6 \mathrm{~m} \
g & =9 \cdot 8 \mathrm{~m} / \mathrm{s}^2 \
	herefore \quad v_0 & =\sqrt{6 \cdot 4 	imes 10^6 	imes 9 \cdot 8} \
& =7.92 	imes 10^3 \mathrm{~m} / \mathrm{s} \
& =7.92 \mathrm{~km} / \mathrm{s}
\end{aligned} \$
The escape velocity of space ship
$ \\begin{aligned}
v_e & =\sqrt{(2 R g)}=\sqrt{2} v_0=\sqrt{2} 	imes 7 \cdot 92 \
& =11 \cdot 20 \mathrm{~km} / \mathrm{s}
\end{aligned}$
$\ 	herefore$ Additional velocity required
$=11 \cdot 20-7 \cdot 92=3 \cdot 28 \mathrm{~km} / \mathrm{s} \$
Therefore, velocity $3.28 \mathrm{~km} / \mathrm{s}$ must be added to orbital velocity to spaceship to overcome the gravitational pull.

Answer

3.28 km/s

Answer

12 km/s

Answer

10 km/s

Answer

40 km/s

Gravitation

A space ship is launched into a circular orbit close to earth’s surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull? (Radius of earth = 6400 km, g = 9.8 m/s^{2})

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