A remote-sensing satellite of earth revolves in a circular orbit at a height of $0.25 \times 10^6\, m$ above the surface of earth. If earth's radius is $6.38 \times 10^6$ m and $g = 9.8 \,ms^{-2}$, then the orbital speed of the satellite is

Question

Two identical particles each of mass $ m $ go around a circle of radius $ a $ under the action of their mutual gravitational attraction. The angular speed of each particle will be:

Answer 1

To determine the orbital speed of the satellite, we follow these steps:

First, understand that the satellite is revolving in a circular orbit. The formula for orbital speed $v$ of a satellite in orbit is given by: v = \sqrt{\frac{GM}{r}} where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the distance from the center of the Earth to the satellite.
From the problem, the height of the satellite above the Earth's surface is 0.25 \times 10^6\, m. The radius of the Earth is 6.38 \times 10^6\, m. Thus, the distance from the center of the Earth to the satellite is: r = 6.38 \times 10^6\, m + 0.25 \times 10^6\, m = 6.63 \times 10^6\, m.
We use the relation between gravitational acceleration $g$ at the Earth's surface and the gravitational constant $G$: \[ g = \frac{GM}{R^2} \] where $R$ is the Earth's radius. We rearrange it to find $GM$: \[ GM = gR^2 \] Plugging in the values, $g = 9.8\, \text{m/s}^2$ and $R = 6.38 \times 10^6\, m$, we have: \[ GM = 9.8 \times (6.38 \times 10^6)^2 \]
Now, substitute the value of $GM$ and $r$ into the orbital speed formula: \[ v = \sqrt{\frac{9.8 \times (6.38 \times 10^6)^2}{6.63 \times 10^6}} \] Calculating this gives us the orbital speed: \[ v \approx 7763\, \text{m/s} \] In km/s, this is approximately 7.76 \, \text{km/s}.
Therefore, the correct option for the orbital speed of the satellite is 7.76 \, km\,s^{-1}.

The correct answer is confirmed as 7.76 \, km\,s^{-1}.

Answer 2

To determine the orbital speed of the satellite, we follow these steps:

First, understand that the satellite is revolving in a circular orbit. The formula for orbital speed $v$ of a satellite in orbit is given by: v = \sqrt{\frac{GM}{r}} where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the distance from the center of the Earth to the satellite.
From the problem, the height of the satellite above the Earth's surface is 0.25 \times 10^6\, m. The radius of the Earth is 6.38 \times 10^6\, m. Thus, the distance from the center of the Earth to the satellite is: r = 6.38 \times 10^6\, m + 0.25 \times 10^6\, m = 6.63 \times 10^6\, m.
We use the relation between gravitational acceleration $g$ at the Earth's surface and the gravitational constant $G$: \[ g = \frac{GM}{R^2} \] where $R$ is the Earth's radius. We rearrange it to find $GM$: \[ GM = gR^2 \] Plugging in the values, $g = 9.8\, \text{m/s}^2$ and $R = 6.38 \times 10^6\, m$, we have: \[ GM = 9.8 \times (6.38 \times 10^6)^2 \]
Now, substitute the value of $GM$ and $r$ into the orbital speed formula: \[ v = \sqrt{\frac{9.8 \times (6.38 \times 10^6)^2}{6.63 \times 10^6}} \] Calculating this gives us the orbital speed: \[ v \approx 7763\, \text{m/s} \] In km/s, this is approximately 7.76 \, \text{km/s}.
Therefore, the correct option for the orbital speed of the satellite is 7.76 \, km\,s^{-1}.

The correct answer is confirmed as 7.76 \, km\,s^{-1}.

A remote-sensing satellite of earth revolves in a circular orbit at a height of $0.25 \times 10^6\, m$ above the surface of earth. If earth's radius is $6.38 \times 10^6$ m and $g = 9.8 \,ms^{-2}$, then the orbital speed of the satellite is

The Correct Option is C

Solution and Explanation

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