2. A satellite in a circular orbit experiences at centripetal acceleration of 11. 89 m / s^2. The tangential speed of the satellite is 7. 76 x 10^3 m / s. What is the altitude of the satellite? (radius of the earth = 6. 38 x 10^6 m)

Answers

Answer 1

The altitude is Altitude ≈ -1,286,502.93 m

How to solve for the altitude

To determine the altitude of the satellite, we can use the centripetal acceleration formula and the equation for tangential speed in circular motion. Here's how we can solve it:

1. Given:

Centripetal acceleration (ac) = 11.89 m/s^2

Tangential speed (v) = 7.76 x 10^3 m/s

Radius of the Earth (r) = 6.38 x 10^6 m

2. Centripetal acceleration (ac) is related to tangential speed (v) and radius (r) by the formula:

ac = v^2 / r

Rearranging the formula to solve for radius (r), we have:

r = v^2 / ac

3. Substitute the given values into the equation:

r = (7.76 x 10^3 m/s)^2 / 11.89 m/s^2

Calculating the right-hand side:

r = 60,537,600 m^2/s^2 / 11.89 m/s^2

r ≈ 5,093,497.07 m

4. The altitude of the satellite is the distance above the Earth's surface, which can be calculated by subtracting the radius of the Earth:

Altitude = r - Radius of Earth

Altitude ≈ 5,093,497.07 m - 6.38 x 10^6 m

Altitude ≈ -1,286,502.93 m

The negative result indicates that the satellite is below the surface of the Earth, which is not possible. Please double-check the given values and ensure that they are accurate for a valid calculation.

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El número sumado que te da 116 y el mismo número si se resta de 18 es 67.

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Llamemos al número desconocido "x". Sabemos que si a un número determinado le sumamos "x", el resultado es 116. Entonces, la ecuación sería: x + y = 116. De igual forma, si a ese mismo número le restamos "x", el resultado es 18: y - x = 18.

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