3. A satellite has a mass of 5890 kg and is in a circular orbit 4.20 x 10³ m above the surface of a planet. The period ofthe orbit is 2.45 hours. The radius of the planet is 4.40 x 10 m. What would be the true weight of the satellite if it wereat rest on the planet's surface? (G= 6.67 x 10−¹¹ N.m²/kg²)X 14373.6 N5TMacBook Air

Name: 3. A satellite has a mass of 5890 kg and is in a circular orbit 4.20
Uploaded: 2024-03-20T06:56:46.000Z
Channel: Bartleby
Description: 3. A satellite has a mass of 5890 kg and is in a circular orbit 4.20 x 10³ m above the surface of a planet. The period ofthe orbit is 2.45 hours. The radius of the planet is 4.40 x 10 m. What would be the true weight of the satellite if it wereat re

Given, mass of satellite , m = 5890 Kg circular orbit of 4.20×105 m above the surface of planet.…

Answered: 3. A satellite has a mass of 5890 kg…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter7: Rotational Motion And Gravitation

Section: Chapter Questions

Problem 2CQ: Suppose an alien civilization has a space station in circular orbit around its home planet. The...

See similar textbooks

Related questions

Concept explainers

Topic Video

Question

3. A satellite has a mass of 5890 kg and is in a circular orbit 4.20 x 10³ m above the surface of a planet. The period of
the orbit is 2.45 hours. The radius of the planet is 4.40 x 10 m. What would be the true weight of the satellite if it were
at rest on the planet's surface? (G= 6.67 x 10−¹¹ N.m²/kg²)
X 14373.6 N
5
T
MacBook Air

Expert Solution

Step 1

Given,

mass of satellite , m = 5890 Kg

circular orbit of $4.20 \times 10^{5} m a b o v e t h e s u r f a c e o f p l a n e t .$

The period of orbit is 2.45 hours.

The radius of planet is $4.40 \times 10^{6} m .$

G = $6.67 \times 10^{- 11} N . m^{2} / k g^{2}$

Total distance of orbit from center of planet = $4.20 \times 10^{5} m + 4.40 \times 10^{6} m = 4.82 \times 10^{6} m$

Let the mass of planet be M.

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions