1. Calculate the velocity of escape of an artificial satellie projected from the earth.Given,mass of earth=5.9*10^24kg,radius of earth=6370km,G=6.67*10^-11Nm^2/kg^2.
 
2.An earth satellite makes a complete circle around the earth in 90 minutes.Assuming the orbit to be circular,calculate the height of the satellite above the earth.Radius of earth=6370km,g=9.80m/s^2.
 
3.Estimate the height above the earth at which the geostationary satellite is moving round the earth.Radius of earth=6400km. Mass of earth=6*10^24.
 
4.Determine the escape velocity of a body from the moon. radius of moon=1.74*10^6m.mass of moon=7.36*10^22kg. G=6.66*10^-11Nm^2/kg^2.
 
5.Jupiter has a mass 318 times that of the earth,and its radius is 11.2 times the radius of the earth.Estimate the escape velocity of a body from the surface of the jupiter,given that the escape velocity from the earth's surface is 11.2km/s.
 
6.An artifical satellite of mass 200 kg revolves round the earth in an orbit of average radius of avaerage radius 6670km. calculate the orbital KE, gravitational potential energy and total energy of the satellite in the orbit.Given mass of earth =6.4*10^34kg, G=6.67*10^-11Nm^2/kg^2
 
7.world's first man made artificial satellite sputnik I was orbiting the earth at a distance of 900km.Calculate its velocity. radius of earth=6370km. g=9.8m/s^2.
 
8.if a satellite is to circle the earth at 1000km. above the surface. with only the attraction of the earth acting  on it, at what speed must it travel? mass of earth=5.9*10^24kg. radius of earth=6370km. G=6.67*10^-11Nm^2/kg^2.
 
9.the moon moves around the earth with a period of 27.3 day's. find the acceleration of the moon towards the center of the earth assuming that the orbit is circular with a radius of 384000km.
 
 
                                                                         sir, please help me to solve this previous year question

Asked by kuriakose7600 | 30th Nov, 2015, 09:49: PM

Expert Answer:

4. 
Given: 
G = 6.67 × 10-11 N m2 kg-2
Radius of the Moon, R = 1.74 × 106 m
Mass of the Moon, M = 7.36 × 1022 kg
If ve is the escape velocity of the body from the Moon, then
begin mathsize 14px style straight v subscript straight e equals square root of fraction numerator 2 GM over denominator straight R end fraction end root equals square root of fraction numerator 2 cross times 6.67 space cross times space 10 to the power of negative 11 end exponent cross times 7.36 space cross times space 10 to the power of 22 over denominator 1.74 space cross times space 10 to the power of 6 end fraction end root equals 2375.4 space straight m space straight s to the power of negative 1 end exponent equals 2.38 space km space straight s to the power of negative 1 end exponent end style
 
5. Let M be the mass and R be the radius of the Earth.
If M' and R' are the mass and radius of the Jupiter, then
M' = 318 M and R' = 11.2 R
Also, the escape velocity on the Earth, ve = 11.2 km s-1
Let ve' be the escape velocity on the Jupiter.
Then,
begin mathsize 14px style straight v subscript straight e to the power of apostrophe over straight v subscript straight e equals square root of fraction numerator 2 GM apostrophe over denominator straight R apostrophe end fraction end root cross times square root of fraction numerator straight R over denominator 2 GM end fraction end root equals square root of fraction numerator straight R over denominator straight R apostrophe end fraction cross times fraction numerator straight M apostrophe over denominator straight M end fraction end root end style
begin mathsize 14px style straight v subscript straight e apostrophe equals straight v subscript straight e square root of fraction numerator straight R over denominator straight R apostrophe end fraction cross times fraction numerator straight M apostrophe over denominator straight M end fraction end root equals 11.2 cross times square root of fraction numerator straight R over denominator 11.2 straight R end fraction cross times fraction numerator 318 straight M over denominator straight M end fraction end root equals 11.2 cross times square root of fraction numerator 318 over denominator 11.2 end fraction end root equals 59.68 space km space straight s to the power of negative 1 end exponent end style
 
Note: Ask the other questions as a separate query.

Answered by Faiza Lambe | 1st Dec, 2015, 08:33: AM

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