Relationship between escape velocity and orbital velocity

Ans. The orbital velocity refers to the velocity that is required by an object to reach towards an orbit around a celestial body such as a star or a planet. Whereas escape velocity is the speed required by an object to leave an orbit.

The equation to determine orbital velocity can be represented as:

v_o = \(\sqrt{\frac{GM}{r}}\)

Here, v refers to the orbital speed or the orbital velocity

G is the Newtons’ gravitational constant,

r refers to the distance from the center of the celestial body that is being orbited at a particular point in time

And M refers to the mass of the body that is being orbited, that could be either the sun or planets.

• Hence, the more mass a body has, (the body that is being orbited), the faster the motion of a satellite has to be in order for the satellite to stay in its orbit.
• The farther the object is from the body, the slower its motion has to be in order to stay in its orbit.

Ans. We know that the mass of the Earth = 5.98 x 10²⁴ kg

The radius of the Earth = 6.38 x 10⁶ m

And G, which refers to the Newton's Gravitational Constant = 6.673 x 10^-11 N m² kg^-2

Now, the escape velocity of the Earth can be calculated by

v_e = \(\sqrt{\frac{2GM}{R}}\)

Now, as we substitute the values known to us in the given equation

v_e = \(\sqrt{\frac{2 (6.673 \times 10^{-11}) (5.98 \times 10^{24})}{6.38 \times 10^{6}}}\)

v_e = \(\sqrt{\frac{7.981 \times 10^{14}}{6.38 \times 10^{6}}}\)

v_e = \(\sqrt{1.251 \times 10^8}\)

v_e = 11184 m/s

v_e = 11.2 km/s

Hence, we can say that the escape velocity required by an object from Earth is 11.2 km/s.

Ans. Given: Mass of the planet (m1) = 8 x mass of the Earth (m2)

Radius of the planet (r1) = 2 x radius of the Earth (r2)

Now, we know that:

Mass of the Earth, m2 = 5.98 x 10²⁴ kg

The radius of the Earth, r2 = 6.38 x 10⁶ m

The value of the Newton gravitational constant, G, is 6.673 × 10^–11 N m² kg^–2

Now, as m1 = 8 x m2

We get the mass of the planet as 8 x 5.98 x 10²⁴ kg

And the radius of the planet as 2 x 6.38 x 10⁶ m

Hence, we can derive the escape velocity by the given formula:

v_e = \(\sqrt{\frac{2GM}{R}}\)

⇒ v_e = \(\sqrt{\frac{2×(6.673×10^{-11})(85.98 ×10^{24})}{2 × 6.38 × 10^6}}\)

⇒ v_e = 22368.96 m s^–1

⇒v_e= 22.36 km s^–1

Hence, the escape velocity of the planet will be 22.36 km/s.

Ans. The escape velocity of an object depends on the following factors:

• The escape velocity primarily depends on the gravitational potential at the point when the body is launched into motion. Now, this potential also depends on the height of the point from the surface of the Earth. Hence, the escape velocity of an object depends on the height from where the object is projected.
• The escape velocity of an object however does not depend on the mass of the body or the direction of the projection of the object.

Ans. The three conditions required for a rocket to transform into a satellite and orbit around the Earth in a circle when it is launched from the surface of the Earth are:

• Firstly, the rocket should be provided with enough vertical velocity that it reaches the height at which it should ideally revolve around the Earth.
• At that certain level of height, the rocket should be provided with a tangential orbital velocity which is given by Vo = √GM/R + h
• At that height of the orbit, the resistance from the air should be almost negligible.

Ans. The mass of the moon is less than the mass of the Earth. Therefore, the escape velocity for the moon is really small. Initially, when the mood had a high temperature, the velocities of the molecules of gas were in the same order as the escape velocity was, therefore the gases escaped the surface of the moon. This is the reason, the moon doesn't have any atmosphere.

Ans. The centripetal force that is required by the satellite to continue its orbital motion is given by the gravitational force of the Earth on the satellite. Therefore, the satellite doesn't need any additional fuel to continue its orbital motion. Even when a satellite is about thousands of miles away from the surface of the Earth, the Earth’s gravity still tugs on it. The circular path of the satellite around the Earth, that is its orbit, is a result of the tug of gravity towards the Earth which is further combined with the momentum from the rocket that is being launched. Satellites that are closer to the planet Earth experience stronger gravity, therefore they have to travel at very high speeds in order to maintain their motion in the orbit and not just fall down back on the surface of the Earth.

Ans. When an object is thrown straight up, the object continues to rise until the gravity’s negative acceleration stops the object’s motion. After that, the object starts on its way to returning back to Earth. As the distance of the object from the center of the Earth increases, the force of gravitation eventually diminishes. Therefore, if we can throw an object upwards with enough of the initial velocity that the decreasing force of the gravity can never slow the object down to a complete stop, then the decreasing velocity of the object can be kept just high enough in order to overcome the pull of the gravity. This initial velocity required by the object to achieve the given condition is referred to as the escape velocity.

Simply, we can define escape velocity as the minimum amount of velocity that is required by an object in order to escape the gravitational field of Earth without falling back to the surface of Earth ever. In order to do so, the object should always have more energy than the gravitational binding energy.

Ans. Given: Radius of the moon = 1.74×10⁶ m

Mass of the moon = 7.35×10²² kg

We know that Newtons Gravitational constant, G = 6.673×10^–11Nm²kg^–2

Therefore, we have:

v_e = \(\sqrt{\frac{2GM}{R}}\)

⇒ v_e = √2 × (6.673 ×10^-11)(7.35 ×10²²) / 1.74 × 10⁶

⇒ v_e = 2374ms^–1

⇒v_e= 2.37 km s^–1

Hence, the escape velocity from the Earth comes out to be 2.37 km s^–1.

Ans. Now, we are given: R = 6.5 × 10⁶ m

M = 5.9722 × 10²⁴ kg

G = 6.67408 × 10^-11 m³ kg^-1 s^-2

Now, the orbital velocity can be calculated by the formula which is denoted as:

V_o = √GM/ R

= √(6.67408 × 10^-11 × 5.9722× 10²⁴) / 6.5 × 10⁶

= √36.68 x 10¹³/ 6.5 x 10⁶

= 7.5 x 10⁹ km/s