according to shell theorems as the body goes near to the core of earth, gravitational force reduces and as it moves away the force increases then why dont the rockets after crossing the earth's atmosphere dont experience any kind of pull due to the gravity ....?
Asked by tejas bhaskar | 6th Dec, 2013, 10:01: PM
Where R = radius of the earth = constant
Thus the potential inside the sphere is independent of positiont .
Since F = we can infer that the shell exerts no force on the particle inside it.
For a solid sphere this means that for a particle, the only gravitational force it feels will be due to the matter closer to center of the sphere (below it). The matter above it (since it is inside its shell) exerts no influence on it.
This is true only for the particle inside the sphere (earth).
But above the surface of the earth distance r (between the rocket and earth) is not constant.
The force is given by F = GMm/r2
From this as the distance increases gravitational force decreases.
So when the rocket flies away from the earth surface, gravitational force on it drops off quadratically with r.
when the rocket escapes earth's atmosphere (with escape velosity), distance r becomes very large (infinite) the gravitational force is negligible ( assumed to be zero).
Answered by Komal Parmar | 9th Dec, 2013, 04:15: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number