QJEEM23Phcb0d: JEE Main 2023 Physics - Class 11 JEEM23Phcb0d Solution: A space ship of mass is launched into a circular orbit close to …

To find the additional velocity required to overcome the gravitational pull and launch the spaceship into orbit, we first need to find the orbital velocity. The formula for orbital velocity (v) is given by:

$v = \sqrt{\frac{GM}{r}}$

where G is the gravitational constant, M is the mass of Earth, and r is the distance from the center of the Earth to the spaceship (which is the sum of the Earth's radius and the altitude of the spaceship's orbit).

In this problem, the spaceship is orbiting close to Earth's surface, so we can approximate r as the Earth's radius. Given that g = 10 m/s² and Earth's radius R = 6400 km, we can relate the gravitational constant G and the mass of Earth M through the formula:

$g = \frac{GM}{R^2}$

Now, we can find the orbital velocity:

$v = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}$

Converting the Earth's radius to meters:

$R = 6400 \times 10^3 \mathrm{~m}$

Plugging in the values:

$v = \sqrt{10 \times 6400 \times 10^3} = 8 \times 10^3 \mathrm{~m/s}$

Now, we need to find the additional velocity required to overcome the gravitational pull. To do this, we can use the formula for escape velocity:

$v\text{escape} = \sqrt{2} \times v\text{orbital}$

Finding the additional velocity required:

$\Delta v = v\text{escape} - v\text{orbital} = (\sqrt{2} - 1) \times v_\text{orbital}$

Plugging in the values:

$\Delta v = (\sqrt{2} - 1) \times 8 \times 10^3 \mathrm{~m/s}$

Converting the velocity to km/s:

$\Delta v = (\sqrt{2} - 1) \times 8 \mathrm{~km/s}$

Thus, the additional velocity required to overcome the gravitational pull and launch the spaceship into orbit is:

$8(\sqrt{2}-1) \mathrm{~km/s}$