Question

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surface, the ship's orbital speed is 5500 m/s . By observing the planet, you determine its radius to be 4.48×106 m. You then land on the surface and, at a place where the ground is level, launch a small projectile with initial speed 14.6 m/s at an angle of 30.8∘ above the horizontal.If resistance due to the planet's atmosphere is negligible, what is the horizontal range of the projectile?

Answer 1

Answer:

$R = 24.3 m$

Explanation:

As we know that the orbital speed is given as

$v = \sqrt{\frac{GM}{R + h}}$

here we know that

v = 5500 m/s

$R = 4.48 \times 10^6 m$

$h = 630 km$

now we have

$5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}$

$M = 2.34 \times 10^24 kg$

now acceleration due to gravity of planet is given as

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

$a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}$

$a = 7.7 m/s^2$

now range of the projectile on the surface of planet is given as

$R = \frac{v^2 sin2\theta}{g}$

$R = \frac{14.6^2 sin(2\times 30.8)}{7.7}$

$R = 24.3 m$

Answer 2

The horizontal range of the projectile is mathematically given as

R = 24.3 m

What is the horizontal range of the projectile?

Question Parameter(s):

at a distance of 630 km above the planet's surface

the ship's orbital speed is 5500 m/s

radius to be 4.48×106 m

a small projectile with initial speed 14.6 m/s

at an angle of 30.8∘ above the horizontal

Generally, the equation for the  orbital speed is mathematically given as

$v = \sqrt{\frac{GM}{R + h}}$

Therefore

$5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}$

M=2.24*10^2kg

Hence

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

$a = \frac{(6.6 *10^{-11})(2.34 * 10^{24})}{(4.48* 10^6)^2}$

a = 7.7 m/s^2

In conclusion, range of the projectile

$R = \frac{14.6^2 sin(2*30.8)}{7.7}$

R = 24.3 m

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