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The radius of Mars (from the center to just above the atmosphere) is 3400 km (3400 × 103 m), and its mass is 6 × 1023 kg. An object is launched straight up from just above the atmosphere of Mars.

a) What initial speed is needed so that when the object is far from Mars its final speed is 3000 m/s? vinitial = _________ m/s

b) What initial speed is needed so that when the object is far from Mars its final speed is 0 m/s? (This is called the "escape speed.") vescape = ______ m/s

Respuesta :

Answer:

Explanation:

Potential energy at the surface

= -GMm / R

= -6.6 x 10⁻¹¹ x 6 x 10²³ m / 3400 x 10³

= -1.1647 x 10⁷ m

If v be required velocity

Total energy at surface

= 1/2 m v²   -1.1647 x 10⁷ m

= energy at farr off = 1/2 m x 3000²

1/2 m v² -  -1.1647 x 10⁷ m = 1/2 m x 9 x 10⁶

1/2 v²= 1.1647 x 10⁷ + .45 x 10⁷

= 1.6147 x 10⁷

v² = 2.3303 x 10⁷

v = 4.827 x 10³ m / s

b ) For final speed zero at infinity

1/2 m v² -  -1.1647 x 10⁷ m = 0

1/2 v² = 1.1647 x 10⁷

v² = 2.3294 x 10⁷

v = 4.826 x 10³ m /s

a. The initial speed is needed should be 4.827 x 10³ m / s.

b. The  initial speed is needed should be 4.826 x 10³ m /s

Calculation of the initial speed:

Since

The potential energy at the surface

= -GMm / R

= -6.6 x 10⁻¹¹ x 6 x 10²³ m / 3400 x 10³

= -1.1647 x 10⁷ m

In the case when v be required velocity

Now

Total energy at surface

= 1/2 m v²   -1.1647 x 10⁷ m

So,

= energy at farr off = 1/2 m x 3000²

1/2 m v² -  -1.1647 x 10⁷ m = 1/2 m x 9 x 10⁶

1/2 v²= 1.1647 x 10⁷ + .45 x 10⁷

= 1.6147 x 10⁷

Now

v² = 2.3303 x 10⁷

v = 4.827 x 10³ m / s

b ) Now the initial speed should be

1/2 m v² -  -1.1647 x 10⁷ m = 0

1/2 v² = 1.1647 x 10⁷

v² = 2.3294 x 10⁷

v = 4.826 x 10³ m /s

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