Planet X rotates in the same manner as the earth, around an azis through its north and south poles, and is perfectly spherical. An astronaut who weighs 943.0 N on the earth weighs 915.0 N at the north pole of Planet X and only 850.0 N at its equator. The distance from the norht pole to the quator is 18,850 km, measured along the surface of Planet X.

1. How long is the day on Planet X?

2. If a 45,000 kg satellite is placed in a cicular orbit 2000 km above the surface of Planet X, what will be its orbital period?

Solution Preview

The difference in weight at the equator and the north pole is due to the centripetal acceleration. Denote the mass of the astronaut by m, the planet's radius by R and the angular rotation speed by omega. Then the decrease in weight as the astronaut moves to the equator is m omega^2 R. This yields the equation:

m omega^2 R = (915.0 - 850.0) N = 65.0 N (1)

The mass m is known because you know that on earth the astronaut weighs 943 N:

m * 9.81 m/s^2 = 943.0 N -->

m = 96.13 Kg (2)

Inserting (2) in (1) gives:

omega^2 R = 0.676 m/s^2 (3)

The distance from the north pole to the equator along the surface is a quarter of an entire circle around the planet from the north pole to the south pole and back, so the distance in terms of R is 2 pi ...

Solution Summary

A detailed solution is given for gravitation, centripetal acceleration, and weight.

... is maximum at the lowest point, the centripetal acceleration will be ... Hence, tangential acceleration is also zero ...Gravitational force exerted by the moon on the ...

... This centripetal force is provided by the force of gravitation between the ...Acceleration due to rotation (centrifugal acceleration) is given by : a = v2/r ...

... What is the horizontal acceleration of the crate ... For the stability of the satellite, gravitational force exerted ... satellite is equal to the centripetal force = mv ...

... The gravitational force experienced by the moon behaves as the centripetal force which produces centripetal acceleration which is ... to hold a 10lb weight next to ...

... it fast enough so that the centrifugal force will be ... and proportional to the object's weight), so you ... that (where g is the gravitational acceleration and µ is ...

... direction, the ball is subject to the gravitational acceleration. ... m3 = 445.99kg So the weight underwater is ... v2 The centripetal acceleration is a = , where r is ...

... rg rg rg g Centripetal Acceleration Centripetal Force ie tan θ = = Gravitational Acceleration Weight Table showing various physical quantities and their ...

... t=0.5 second under the gravitational acceleration g. During ... At the bottom point, Total force = Centrifugal (in the ... will get total ie resultant acceleration v2 a ...

... F = GMm/R2 where G = Universal gravitational constant, M ... at a constant speed is undergoing acceleration and has a ... also act towards the centre (centripetal force ...

... Where Me is earth's mass, G is the gravitational constant and ...weight of the stick (5N) and the weight of the ... The centripetal acceleration that they will feel is ...