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Q8.22:
As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth = $6.0 \times 10^{24}$ kg, radius = 6400 km.
Solution :
Step 1: Identification of Given Variables
To determine the gravitational potential $V$ at the site of a geostationary satellite, we define the parameters provided:
- Mass of Earth ($M$): $6.0 \times 10^{24}\ \text{kg}$
- Radius of Earth ($R$): $6400\ \text{km} = 6.4 \times 10^6\ \text{m}$
- Height of satellite ($h$): $36,000\ \text{km} = 3.6 \times 10^7\ \text{m}$
- Gravitational constant ($G$): $6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$
Step 2: Defining the Orbital Radius
The gravitational potential is dependent on the distance from the center of the Earth. The distance $r$ from the center to the satellite is the sum of Earth's radius and the altitude:
$r = R + h$
$r = 6.4 \times 10^6\ \text{m} + 36 \times 10^6\ \text{m}$
$r = 4.24 \times 10^7\ \text{m}$ [By definition of radial distance in a central force field]
Step 3: Formulating the Potential Equation
The gravitational potential $V$ at a point at distance $r$ from a point mass (or spherical mass) $M$ is defined as the work done per unit mass to bring a test mass from infinity to that point [Per the definition of Gravitational Potential Energy $U = -GMm/r$]:
$V = -\frac{GM}{r}$
Step 4: Calculation
Substituting the values into the derived formula:
$V = -\frac{(6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2) \times (6.0 \times 10^{24}\ \text{kg})}{4.24 \times 10^7\ \text{m}}$
$V = -\frac{40.02 \times 10^{13}}{4.24 \times 10^7}\ \text{J/kg}$
$V \approx -9.438 \times 10^6\ \text{J/kg}$
Step 5: Analysis of Results
The negative sign indicates that the gravitational force is attractive and that the potential is lower than the potential at infinity (which is defined as zero) [Per the convention of gravitational field theory].
Final Solution: The gravitational potential at the site of the geostationary satellite is approximately $-9.44 \times 10^6\ \text{J/kg}$.
More Questions from Class 11 Physics Gravitation Additional Exercises
- Q8.23: A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity ? (mass of the sun = $2 \times 10^{30}$ kg).
- Q8.24: A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system ? Mass of the space ship = 1000 kg; mass of the sun = $2 \times 10^{30}$ kg; mass of mars = $6.4 \times 10^{23}$ kg; radius of mars = 3395 km; radius of the orbit of mars = $2.28 \times 10^8$ km; G = $6.67 \times 10^{-11}$ N m$^2$ kg$^{–2}$.
- Q8.25: A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s$^{–1}$. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it ? Mass of mars = $6.4 \times 10^{23}$ kg; radius of mars = 3395 km; G = $6.67 \times 10^{-11}$ N m$^2$ kg$^{–2}$.
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