Question

What is the station's orbital speed? the radius of earth is 6.37×106m, its mass is 5.98×1024kg.

Answer 1

Answer:

$v_o=2503.08\frac{m}{s}$

Explanation:

Orbital velocity is the speed that a body that orbits around another body must have, for its orbit to be stable. For orbits with small eccentricity and when one of the masses is almost negligible compared to the other mass, like in this case, the orbital speed is given by:

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

Where M is the greater mass around which this negligible body is orbiting, r is the radius of the greater mass and G is the universal gravitational constant. So:

$v_o=\sqrt{\frac{6.674*10^{-11}\frac{m^3}{kg\cdot s^2}(5.98*10^{24}kg)}{6.37*10^6m}}\\v_o=2503.08\frac{m}{s}$

Answer 2

The station's orbital speed is about 7.69 × 10³ m/s

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Further explanation

Complete Question:

The International Space Station is in a 230-mile-high orbit.

What is the station's orbital speed? The radius of Earth is 6.37×10⁶ m, its mass is 5.98×10²⁴ kg

Given:

height of the station = h = 230 miles = 3.70 × 10⁵ m

mass of the earth = M = 5.98 × 10²⁴ kg

radius of the earth = R = 6.37 × 10⁶ m

Unknown:

Orbital Speed of the station = v = ?

Solution:

We will use this following formula to find the orbital speed:

$F = ma$

$G \frac{ Mm}{(R+h)^2}=m v^2 \div (R+h)$

$G \frac{ M}{R+h} = v^2$

$v = \sqrt{ G \frac{M}{R+h}}$

$v = \sqrt{ 6.67 \times 10^{-11} \frac{5.98 \times 10^{24}}{6.37 \times 10^6 + 3.70 \times 10^5}}$

$v = 7.69 \times 10^3 \texttt{ m/s}$

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Learn more

• Impacts of Gravity : https://brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
• The Acceleration Due To Gravity : https://brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant