Home
Class 14
GENERAL KNOWLEDGE
The acceleration due to gravity at the e...

The acceleration due to gravity at the equator

A

is less than that at the poles

B

is greater than that at the poles

C

is equal to that at the poles

D

does not depend on the earth's centripetal acceleration

Text Solution

Verified by Experts

The correct Answer is:
A

is less than that at the poles
Promotional Banner

Similar Questions

Explore conceptually related problems

Calculate that imaginary angular velocity of the Earth for which effective acceleration due to gravity at the equator becomes zero. In this condition, find the length (in hours) of a day? Radius of Earth = 6400 km. g = 10 ms^(-2) .

In order to make the effective acceleration due to gravity at the equator to zero, the numerical value of the angular velocity of rotation of the earth should be [Radius of the earth=6400 km, g=10m/s^(2) ]

Find the imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero. (take g = 10 m//s^(2) for the acceleration due to gravity, if the earth were at rest and radius of earth equal to 6400 km and phi = 60^(@) )

The imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero is equal to [Take g = 10 m//s^(2) for the acceleration due to gravity if the earth were at rest and radius of earth equal to 6400 km.]

Velocity of a body at the equator is v .The escape velocity of the body at the poles if the value of acceleration due to gravity at the equator is (1)/(3) of the value at the poles is (Average radius of the earth is R )

A uniform spherical planet is rotating about its axis. The speed of a point on its equator is v and the effective acceleration due to gravity on the equator is one third its value at the poles. Calculate the escape velocity for a particle at the pole of the planet. Give your answer in multiple of v.

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V . Due to the rotation of planet about its axis the acceleration due to gravity g at equator is 1//2 of g at poles. The escape velocity of a particle on the planet in terms of V .

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is 7.5 kms^(-1) . Due to the rotation of the planet about its axis, the acceleration due to gravity g at equator is 1//2 of g at poles. What is the escape velocity ("in km s"^(-1)) of a particle on the planet from the pole of the planet?

What is the value of the acceleration due to gravity at the centre of the earth?

The acceleration due to gravity near the surface of moon is-

Recommended Questions
  1. The acceleration due to gravity at the equator

    Text Solution

    |

  2. विभिन्नताए क्या है ? विभिन्नता के कारण सविस्तार लिखिए !

    Text Solution

    |

  3. जीवो की अप्राकृतिक विलुप्ति के कारण लिखिए!

    Text Solution

    |

  4. 450 वाट की एक विघुत प्रेस 225 वोल्ट पर कार्य करने के लिए बनाई गई हैं ।...

    Text Solution

    |

  5. छोटे दण्ड-चुम्बक के कारण अनुप्रस्थ ( निरक्षीय ) स्थिति में किसी बिन्दु...

    Text Solution

    |

  6. चित्रानुसार एक दण्ड चुम्बक मुक्त रूप से एक कुण्डली के बीच से होकर गिरत...

    Text Solution

    |

  7. चुम्बकीय द्विध्रुव के कारण अक्षीय स्थिति में किसी बिन्दु पर चुम्बकीय क...

    Text Solution

    |

  8. 2.0 मिमी व्यास के लम्बे ताँबे के तार में 10 A की धारा है। इस धारा के ...

    Text Solution

    |

  9. किसी स्थान पर पृथ्वी के चुम्बकीय क्षेत्र का क्षैतिज तथा ऊर्ध्वाधर घटक ...

    Text Solution

    |