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Where the gravitational field strength is maximum?

value of acceleration due to gravity is maximum at the surface of earth. So the gravitational field strength. as g'=g(1-d/R) at surface d=R so d=R so g'=g at earth's centre g=0. Its value decrease with decrease or increase in height. as: g'=g(1-2h/R) ......for height h and g'=g(1-d/R) .....for depth d


What is the approximate value of g at an altitude above Earth equal to 1.47 times Earth diameter?

The formula to calculate the acceleration due to gravity at a certain altitude h above Earth's surface is g' = g * (R / (R + h))^2, where g is the acceleration due to gravity at the Earth's surface (approximately 9.81 m/s^2), R is the radius of the Earth (approximately 6371 km), and h is the altitude. In this case, h = 1.47 * R, so substituting the values into the formula gives g' = 5.64 m/s^2.


How does the mass of the plnnet affect its gravity?

According to newtons formula; force F=G*m1*m2/(r^2) ,for 2 bodies facing each others gravitational pull When divided both sides by m1,so gravitational acceleration g=m2*g/(r^2) so g is directly proportional to mass of the body....


What is the gravitational attraction between mars moons?

It greatly depends upon their distance to one another at the time. However, the universal law of gravitational attraction applies: F = G * ((m1*m2)/r) where m1 is the mass of moon 1 (kg) m2 is the mass of moon 2 (kg) r is the distance (m) G is the gravitational constant F is the force of attraction.


A 124 kg satellite experiences a gravitational force by the Earth of 690 N what is the radius of the satellite's orbit in kilometers?

F= G (m1m2)/(r2) F= the gravitational force G= gravitational constant m1= mass of the first object (the satellite) m2= mass of the second object (earth) r= the radius Plug in the values and solve for r: 690 N= 6.67 X 10-11 ((124kg) X (5.98 X 1024)/(r2) 690r2= 6.67 X10-11 (7.41 X 1026) 690r2= 4.94 X 1016 r2= (4.94 X 1016)/(690) r= square root of (7.16 x 1013) r= 8.46 x 106 m, or 846,000 Km