When the space shuttle (mass of the shuttle = 2 x10^6 kg) was at the outer limit of Earth's atmosphere (distance = 350 km above Earth's surface), astronauts measured the gravitational force from Earth to be equal to 1.7 x 10^7 newtons. Assuming that both Earth and the shuttle are point masses, calculate the gravitational force of Earth on the shuttle when the shuttle was at the limit of the mesosphere, i.e., at 90 km from Earth's surface. Also, what is the gravitational force the shuttle exerts on Earth at this same point?

F = G Mearth Mshuttle /r^2

G , Mearth and Mshuttle are the same

so
F r^2 = same
F1 r1^2 = F2 r2^2
F2 = F1 r1^2/r2^2

F1 = 1.7*10^7
r1 = 350*10^3 + 6.38*10^6 = 6.73*10^6 meters
r2 = 90*10^3 + 6.38*10^6 = 6.47*10^6 meters
so
F2 = 1.7*10^7 (6.73/6.47)^2
= 1.7 *10^7 * 1.082
= 1.84 * 10^7 Newtons

Newton's third law applies o the second part of the question.

To calculate the gravitational force between two point masses, we can use the equation:

F = G * (m1 * m2) / r^2

Where:
F is the gravitational force
G is the gravitational constant (approximately 6.67 x 10^-11 N m^2/kg^2)
m1 and m2 are the masses of the two point masses
r is the distance between the two point masses

In this case, the mass of the Earth (m1) is much greater than the mass of the shuttle (m2), so we can consider the Earth's mass to be constant.

1. To calculate the gravitational force of Earth on the shuttle at 350 km above the Earth's surface (distance = 350,000 meters):

F1 = G * (m1 * m2) / r^2

= (6.67 x 10^-11 N m^2/kg^2) * (2 x 10^6 kg) * (5.97 x 10^24 kg) / (350,000 m + Earth's radius)^2

Note: Earth's mass is approximately 5.97 x 10^24 kg.

2. To calculate the gravitational force of Earth on the shuttle at 90 km above the Earth's surface (distance = 90,000 meters):

F2 = G * (m1 * m2) / r^2

= (6.67 x 10^-11 N m^2/kg^2) * (2 x 10^6 kg) * (5.97 x 10^24 kg) / (90,000 m + Earth's radius)^2

3. The gravitational force the shuttle exerts on Earth at the same point is equal in magnitude but opposite in direction to the gravitational force of Earth on the shuttle. Therefore, the gravitational force the shuttle exerts on Earth would be equal to the gravitational force of Earth on the shuttle at that point:

F3 = F2 (gravitational force of Earth on the shuttle at 90 km above Earth's surface)

= G * (m1 * m2) / r^2

= (6.67 x 10^-11 N m^2/kg^2) * (2 x 10^6 kg) * (5.97 x 10^24 kg) / (90,000 m + Earth's radius)^2

These calculations should give you the gravitational forces between the Earth and the shuttle at the given distances above the Earth's surface.

To calculate the gravitational force between two point masses, we can use Newton's law of universal gravitation. The formula is given as:

F = (G * m1 * m2) / r^2

Where:
F is the gravitational force
G is the gravitational constant (6.67430 × 10^-11 Nm^2/kg^2)
m1 and m2 are the masses of the two objects
r is the distance between the centers of the two objects

In this case, we have the mass of the Earth (m1) and the mass of the shuttle (m2) and want to calculate the gravitational force.

Given:
Mass of Earth (m1) = 5.972 × 10^24 kg
Mass of Shuttle (m2) = 2 × 10^6 kg
Distance from Earth's surface (r1) = 350 km + Earth's radius (6,371 km)
Distance from Earth's surface (r2) = 90 km + Earth's radius (6,371 km)

Step 1: Calculating the gravitational force from Earth to the shuttle at 350 km distance from Earth's surface.
F1 = (G * m1 * m2) / r1^2
Substituting the values,
F1 = (6.67430 × 10^-11 Nm^2/kg^2) * (5.972 × 10^24 kg) * (2 × 10^6 kg) / (3,721,000 m)^2
Calculating the value of F1 gives us:
F1 = 1.786 × 10^7 N

Step 2: Calculating the gravitational force from Earth to the shuttle at 90 km distance from Earth's surface.
F2 = (G * m1 * m2) / r2^2
Substituting the values,
F2 = (6.67430 × 10^-11 Nm^2/kg^2) * (5.972 × 10^24 kg) * (2 × 10^6 kg) / (6,461,000 m)^2
Calculating the value of F2 gives us:
F2 = 1.758 × 10^7 N

The gravitational force of Earth on the shuttle at the limit of the mesosphere (90 km from Earth's surface) is approximately 1.758 × 10^7 N.

To calculate the gravitational force the shuttle exerts on Earth at the same point, we know that the gravitational force is mutual or equal in magnitude but opposite in direction. Therefore, the gravitational force that the shuttle exerts on Earth at the same point is also approximately 1.758 × 10^7 N.