Question

5. A small asteroid with a mass of 2.05 × 10° kg is pulled into a circular orbit around Earth. The distance from the asteroid to Earth's center is 7378 km. If the gravitational force needed to keep the asteroid in orbit has a magnitude of 3.00 × 10° N, what is the asteroid's tangential speed?

Answer 1

The tangential speed of the asteroid in circular orbit around Earth is approximately 7.55 km/s.

Step-by-step explanation:

To find the tangential speed of the asteroid in circular orbit around Earth, we need to use the formula:

v = √(GM/r)

Where v is the tangential speed, G is the gravitational constant (6.67 × 10-11 Nm2/kg2), M is the mass of Earth (5.97 × 1024 kg), and r is the distance from the asteroid to Earth's center (7378 km = 7.378 × 106 m).

Plugging in these values into the formula, we find:

v = √((6.67 × 10-11 Nm2/kg2)(5.97 × 1024 kg)/(7.378 × 106 m))

Simplifying this expression gives:

v ≈ 7.55 km/s

Therefore, the asteroid's tangential speed is approximately 7.55 km/s.